ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben...

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R M M ROMANIAN MATHEMATICAL MAGAZINE Founding Editor Founding Editor DANIEL SITARU DANIEL SITARU Available online Available online www.ssmrmh.ro www.ssmrmh.ro ISSN-L 2501-0099 ISSN-L 2501-0099 Number 25 Number 25 SUMMER 2022 SUMMER 2022 SOLUTIONS SOLUTIONS

Transcript of ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben...

Page 1: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

R M MROMANIAN MATHEMATICAL MAGAZINE

Founding EditorFounding EditorDANIEL SITARUDANIEL SITARU

Available onlineAvailable onlinewww.ssmrmh.rowww.ssmrmh.ro

ISSN-L 2501-0099ISSN-L 2501-0099

Number 25Number 25 SUMMER 2022SUMMER 2022

SOLUTIONSSOLUTIONS

Page 2: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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Proposed by

Daniel Sitaru-Romania

George Apostolopoulos-Messolonghi-Greece

Marian Ursărescu-Romania

D.M. Bătineţu-Giurgiu - Romania

Neculai Stanciu-Romania

Florică Anastase-Romania

Cristian Miu-Romania

Vasile Mircea Popa – Romania

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Solutions by

Daniel Sitaru-Romania, Asmat Qatea-Afghanistan, Amrit Awasthi-India,

Abdul Aziz-Semarang-Indonesia,Daniel Văcaru-Romania,George

Apostolopoulos-Messolonghi-Greece,Mohamed Amine Ben Ajiba-Tanger-

Morocco,Angel Plaza-Spain,Ravi Prakash-New Delhi-India,Marian Ursărescu-

Romania,Daoudi Abdessattar-Tunisia,D.M. Bătineţu-Giurgiu –

Romania,Neculai Stanciu-Romania,George Florin Șerban-Romania, Samir

Cabiyev-Azerbaijan,Nikos Ntorvas-Greece,Florică Anastase-Romania,Aggeliki

Papaspyropoulos-Greece,Cristian Miu-Romania,Mikael Bernardo-

Mozambique,Timson Azeez Folorunsho-Lagos-Nigeria,Rana Ranino-Setif-

Algerie,Samar Das-India,Akerele Segun-Lagos-Nigeria,Adrian Popa-

Romania,Kaushik Mahanta-Assam-India,Felix Marin-Romania,Jaihon

Obaidullah-Afghanistan,Remus Florin Stanca-Romania,Kamel Gandouli

Rezgui-Tunisia,Ruxandra Daniela Tonilă-Romania,Vasile Mircea Popa –

Romania,Serlea Kabay-Liberia,Ajentunmobi Abdulquyyom-Nigeria,

Mohammad Rostami-Afghanistan

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JP.361 Find �, �, � > 0 such that:

���� + ��� + ��� = ���(� + � + �)

�� + �� + �� = ��

Proposed by Daniel Sitaru-Romania

Solution 1 by proposer

��� + ��� + ��� = ��

���� +

���� +

����� + (

����� +

���� +

����� +

+ ��

���� +

���� +

����� ≥

������ ⋅

��(���)�(���)(���)��

+

+� ⋅�

��(���)�(���)(���)��

+ � ⋅�

��(���)�(���)(���)��

=

= ���������+ ���������

+ ���������=

= ���� + ���� + ���� = ���(� + � + �)

Equality holds for ��� = ��� = ��� ⇔ � = � = �.

�� + �� + �� = �� ⇒ ��� = �� ⇒ � = �.

Solution: � = � = � = �.

Solution 2 by AsmatQatea-Afghanistan

��� + ��� + ��� = ���(� + � + �)|: (���)

→��

�+

��

�+

��

�= � + � + �

��

�+

��

�+

��

�≥⏞

���������

� + � + �

Equality holds when � = � = � → �� + �� + +�� = �� → � = �.

So, � = � = � = �.

Solution 3 by AmritAwasthi-India

� ���

���

= � ����

���

⇔ � ���(� − �)

���

= �

The left summand becomes zero when � = � = �. Let � = � = � = �.

Therefore, putting in the next equation we get,

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�� + �� + �� = �� → �� + �� + �� = �� →

��� = � → � = �. So, � = � = � = �.

Solution 4 by Abdul Aziz-Semarang-Indonesia

��� + ���� + ���� ≥⏞�����

�����

��� + ���� + ���� ≥⏞�����

�����

��� + ���� + ���� ≥⏞�����

�����

Thus,

��� + ��� + ��� ≥ ���(� + � + �)

Since: ��� + ��� + ��� = ���(� + � + �), then: � = � = �.

Since, �� + �� + �� = ��, we have:

� = � = � = �.

Solution 5 by Daniel Văcaru-Romania

���� + ��� + ��� = ���(� + � + �); (�)

�� + �� + �� = ��; (�)

We have: � + � + � =⏞(�)

�����������

���=

��

�+

��

�+

��

�≥⏞

���������(�����)�

�����= � + � + �

� = � = �. From (�), we obtain � = � = � = �.

JP.362. Let ��� be a triangle with inradius �, and circumradius �. Equilateral

triangles with ��, ��and ��, are drawn externally to triangle ���. Let �, �

and � be the centroids of the equilateral triangles, respectively. Prove that:

��

�≤

[���]

[���]≤ �

���

Proposed by George Apostolopoulos-Messolonghi-Greece

Solution 1 by proposer

Let �, �, � be the lengths of sides ��, ��, ��, respectively. Then, we have:

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�� =�

�⋅

�√�

�=

�√�

�, �� =

�⋅

�√�

�=

�√�

We use Law of cosines on ∆���:

��� = ��� + ��� − ��� ⋅ �����(∠���) =��

�+

��

�−

������(�����° + � + ��°)

=�� + ��

�−

���(�����°���� − �����°����) =

=�� + ��

�−

��� �

�⋅

�� + �� − ��

���−

√�

�⋅

��� =

=�� + �� + ��

�+

√�

�⋅

���

��

So, ��� =��������

�+

�√�

�⋅ [���].Similarly, we have:

��� =��������

�+

�√�

�⋅ [���]and ��� =

��������

�+

�√�

�⋅ [���].

Namely, �� = �� = ��.So∆��� is equilateral. Now,

[���] =��� ⋅ √�

�=

√�

��(�� + �� + ��) +

�[���]

We know that: �� + �� + �� ≤ ���and �√�[���] ≤ �� + �� + �� ≤ ���.

Namely, [���] ≤√�

��(���) +

��

�√�

�� =

�√�

���. So,

[���]

[���]≤

�√�� ��

��≤

�√�� ��

�(�√��)=

��

���= �

���

Also, we have: [���] =√�

��(�� + �� + ��) +

�[���] ≥

√�

��(����) +

�(��).

We know that: [���] ≥�√�

��� +

����√��� = �√���. Namely,

[���]

[���]≥

�√���

��≥

�√���

�√�� �

=��

Therefore,

��

�≤

[���]

[���]≤ �

���

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Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco

��� � �� ��� ������������ �� ���� ��� � ��� �������� �� ��.

��� ������� �� ����������� ��������� ������� ������� ���

��������� �� ��, ��, �� ��������� �� �.

�� ���� ∶ �� =�

����

�=

√�

�� ��� �� = � ��� � → ��

= � ��� � +√�

�� (��� �������)

→ [���] = �[���] =�

�� ��. ��. ��� ���

=�

�� �� ��� � +

√�

��� �� ��� � +

√�

��� ���(� − �) =

=��

�� ��� � . ��� � . ��� � +

√��

���(� ��� � + � ��� �) ��� � +

��� �� ��� �

���, � ��� � . ��� � . ��� � = �� ��� �� �� ��� �� = �� ��� �� �� ��� ��

= � ��� � =�

���.

� ��� � + � ��� � =�� + �� − ��

��+

�� + �� − ��

��= � →

�(� ��� � + � ��� �) ��� � =�

��� ��.

��� ∶ � �� ��� � =�

��� ��� =

�. ���

��= �� →

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[���] =��

�.

���+

√��

��.

��� �� +

��. �� =

�� +

√�

��� ��.

���, [���] ≥⏞������� �����������

�� +

√�

��. �√�� = � →

[���]

[���]≥ � ≥⏞

�����

��

� (�).

��� [���] ≤⏞������� �

�� +

√�

��. ��� →

[���]

[���]≤

�+

�√���

���≤

≤⏞����������

�+

�√���

���√���� ≤⏞

�����

��

���

+�

��

���

= ��

���

→[���]

[���]≤ �

���

(�)

(�), (�) → ��

�≤

[���]

[���]≤ �

���

JP.363. Let �, �, � be positive real numbers with �� + �� + �� = ��.

Prove that:

��

√�� + �+

��

√�� + �+

��

√�� + �≥ ��

Proposed by George Apostolopoulos-Messolonghi-Greece

Solution 1 by proposer

We have: ��(� − �)� ≥ �, with equality for � = �. So, ��(�� − �� + �) ≥ � ⇔

�� + ��� + � − ��� − � = (�� + �)� − �(�� + �) ≥ � ⇔ �(�� + �) ≤ (�� + �)� ⇒

�√�� + � ≤ �� + �. Now, �

������≥

����⇔

��

������≥

��

����.

Similarly,

��

�√�� + �≥

��

�� + �;

��

�√�� + �≥

��

�� + �

Adding up these inequalities, we have:

��

��

√�� + �+

��

√�� + �+

��

√�� + �� ≥

��

�� + �+

��

�� + �+

��

�� + �

Applying Cauchy-Schwarz inequality, we get:

��

√�� + �+

��

√�� + �+

��

√�� + �≥ � ⋅

(�� + �� + ��)�

(�� + �� + ��) + �=

� ⋅ ���

�� + � = ��

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Equality holds when � = � = � = �.

Solution 2 by Angel Plaza-Spain

By Jensen’s inequality, by doing �� = �, �� = �, �� = �, the inequality becomes:

��

���� + �

+��

���� + �

+��

���� + �

≥ ��

For �, �, � positive real numbers with � + � + � = ��.

Function �(�) =��

��

����

is convex for � ∈ (�, ��) because

���(�) =���

�� + ��� + ��

�� ���� + ��

��

> �

Therefore, by Jensen’s inequality, we get:

��

���� + �

+��

���� + �

+��

���� + �

≥ � ⋅��

���� + �

= ��

JP.364 If �, �, � > 0 then:

��� − ��√� + �� + ��� − ��√� + �� = ��� + �� −���√� − √��

�⇔

�√�

�+

�=

√� + √�

Proposed by Daniel Sitaru – Romania

Solution 1 by proposer

Denote �� = �, �� = �; �� = �

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������ � = ��°, ������� = ��°, ������ � = ��°

� ∈ �������� �

�� = ��� + �� − ��� ��� ��° = ��� + �� − ��� ⋅√�

�= ��� + �� − ��√�

�� = ��� + �� − ��� ��� ��° = ��� + �� − ��� ⋅√�

�= ��� + �� − ��√�

�� = ��� + �� − ��� ��� ��° = ��� + �� − ���(��� ��° ��� ��° − ��� ��° ��� ��°) =

= ��� + �� − ��� �√�

�⋅

√�

�−

√�

�⋅

�� = ��� + �� −

���√� − √��

��� − ��√� + �� + ��� − ��√� + �� = ��� + �� −���√� − √��

�⇔

⇔ �� + �� = �� ⇔ �, �, � – colineare

⇔ [���] + [���] = [���] ⇔

⇔�

��� ��� ��° +

��� ��� ��° =

��� ��� ��°

⇔ �� ��� ��° + �� ��� ��° = �� ���(��° + ��°)

⇔ �� ⋅�

�+ �� ⋅

√�

�= ��(��� ��° ��� ��° + ��� ��° ��� ��°)

⇔��

�+

��√�

�= �� �

�⋅

√�

�+

√�

�⋅

√�

�� ⇔ �� + ��√� =

���√� + √��

⇔ ��� + �√��� = ���√� + √�� ⇔�

�+

�√�

�=

√� + √�

Solution 2 by Ravi Prakash-New Delhi-India

�� − ��√� + �� = �� − � ����

��� + �� = �� ���

�− ��

+ �� �����

�=

= ��� ����

�− �� + � ���

���

→ ��� − ��√� + �� = ��� ����

�− �� + ����

�� ���

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Next,

��� − ��√� + �� = ��� − ����

��� + ����

�� ���

�� + �� −√� − √�

��� = �� + �� − � ����

����

�− ���

����

�� �� =

= �� ����

�− � ���

��

+ �� ����

�+ � ���

��

=

= ��� ����

�− � ���

�� + �� ���

�+ � ���

�� ��

→ ��� + �� −�

��√� − √���� = ��� ���

�− � ���

�� + �� ���

�+ � ���

�� ��

Now,

��� − ��√� + �� + ��� − ��√� + �� = ��� + �� −�

��√� − √����

⇔ ��� ����

�− �� + ����

�� ��� + ��� − ���

��� + ����

�� ��� =

= ��� ����

�− � ���

�� + �� ���

�+ � ���

�� �� ⇔

� ����� − �

� − � �����

=� ���

��

� �����

⇔ �� �����

����

�� − �� ���

�= �� ���

�− �� ���

����

⇔ �� �√�

�√�+

�√�� =

��

√�+

��

��

⇔� + √�

�√�⋅

�=

�√�+

��⇔

√� + √�

�=

√�

�+

Therefore,

�√�

�+

�=

√� + √�

JP365. If in ���� exists the relationship:

��=

�+

�� then prove that �� ≥ ��.

Proposed by Marian Ursărescu-Romania

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Solution 1 by proposer

��=

�+

��⇔

���� + �

�����

=� + � − �

�⇔

��(� + �)

��� ⋅ �����

= �� − � ⇔

��(� + �)

��� ⋅ ����� ���

��

⋅ ����

�= � + � ⇔

��

�������

⋅ ����

�= �

��

��⋅ ���

�= � ⇒ ���

�=

�⇒

�= ��° ⇒ � = ��°

�� = �� ⋅ ���� = � ⇒ � ≥ ��(�����)

Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco

��=

�+

��↔

�(� + �)

��. �����

=�

��� +

� − �

�� =

� + �

��↔ ��� = ��. ���

�=

����

�. ���

↔ � = �� ����

�↔ �� ���

����

�= �� ���

�↔ ���

�=

�.

���, �� ���� ���� ∶ �� = ��|��� �| = �� �� − � �����

�� = �� �� − �.

�� = � ≥⏞

�����

��

���������, �� ≥ ��.

JP.366. In acute ���� the following relationship holds:

���� + √�������� + √�������������

< �

Proposed by Marian Ursărescu-Romania

Solution by proposer

���� + √�������� + √�������������

=

= ���� +�

�√���� ⋅ ����� +

�√���� ⋅ ����� ⋅ �������

≤ ���� +�

�(���� + �����) +

��(���� + ����� + ������) =

=�

��(������ + ����� + ������ + ���� + ����� + ������) =

=��

��(���� + ���� + ����) =

�(���� + ���� + ����); (�)

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���� + ���� + ���� ≤�

�; (�)

From (1),(2) it follows that:

���� + √�������� + √�������������

<�

�⋅

�< �

Equality holds if and only if triangle is equilateral.

JP.367. �, �, � ∈ ℂ∗ −different in pairs, �(�), �(�), �(�); |�| = |�| = |�| = �.

If

(��)� + (��)� + (��)� = �(���)�, then ���� is equilateral.

Proposed by Marian Ursărescu-Romania

Solution 1 by proposer

�(�), �(�), �(�); |�| = |�| = |�| = �; ���� ⊂ �(�, �) ⇒ |�|� = � ⇒

� ⋅ �� = �; � ⋅ �� = �; � ⋅ �� = �

���������

+ ���������

+ (������)� = ������������

��⋅

��+

��⋅

��+

��⋅

��=

������⇔ �� + �� + �� = ���� ⇔

�� + �� + �� − ���� = � ⇔

(� + � + �)(�� + �� + �� − �� − �� − ��) = �

If � + � + � = � ⇒ �� = �� ⇒ � = � ⇒ ���� −equilateral.

If �� + �� + �� = �� + �� + �� ⇒ ���� −equilateral.

Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco

|�| = |�| = |�| = � → ���� ∈ �(�, � = �).

(∗) ↔ �����

+ �����

+ (��)� = �������

↔�

(��)�+

(��)�+

(��)�=

(���)�

↔ �� + �� + �� = ���� ↔ (� + � + �)(�� + �� + �� − �� − �� − ��) = �

↔ � + � + � = � �� �� + �� + �� − �� − �� − �� = �

�� � + � + � = � → �� = �� → � ≡ � → ���� �� �����������.

�� �� + �� + �� − �� − �� − �� = � ↔� − �

� − �=

� − �

� − �

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↔ ∆��� ~ ∆��� → ���� �� �����������.

���������, (��)� + (��)� + (��)� = �(���)� → ���� �� �����������.

Solution 3 by Daoudi Abdessattar-Tunisia

�(�), �(�), �(�), (� − �)(� − �)(� − �) ≠ � ��� |�| = |�| = |�| = �

������� ����: (��)� + (��)� + (��)� = �(���)�; (�)

(�) ⇔��

��+

��

��+

��

��= � ⇔ � ��(������)

���

= �, (� = ���, � = ���, � = ���, �, �, �

∈ [−�, �]

⇔ �∑���(� + � − ��) = �

∑���(� + � − ��) = �⇔ � + � = �� ⇔ �� = ��, �� = ��, �� = ��

�� = �� → ��� = (��)� = �� → �� = �� ��� ���������, �� = ��.

�, �, � ��� ����� �� �������� �� = ���, −� < � < �

→ {�, �, �} ∈ ����� , �

���

����

� , ����

����

� � → �� = �� = ��.

JP.368. �, �, � ∈ ℂ∗ −different in pairs, �(�), �(�), �(�); |�| = |�| = |�| = �.

If |� − �| ��

�+

�� + |� − �| �

�+

�� + |� − �| �

�+

�� = � , then ���� is

equilateral.

Proposed by Marian Ursărescu-Romania

Solution by proposer

�(�), �(�), �(�); |�| = |�| = |�| = �; ���� ⊂ �(�, �)

|�|� = � ⇒ � ⋅ �� = � ⇒ � ⋅ �� = � ⋅ �� = � ⋅ �� = �.

|� − �| + |� − �|

�+

|� − �| + |� − �|

�+

|� − �| + |� − �|

�= � ⇔

|� − �| + |� − �|

�+

|� − �| + |� − �|

�+

|� − �| + |� − �|

������������������������������������������������������������������������= �

(� + �)� + (� + �)� + (� + �)� = �

(� + � + � − �)� + (� + � + � − �)� + (� + � + � − �)� = � ⇔

(� + � + �)(� + � + �) = �� + �� + �� ⇔

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� + � + � =��������

�����⇔ � = � ⇒ ���� −equilateral.

Solution 2 by Mohamed Amine-Ben Ajiba-Tanger-Morocco

|�| = |�| = |�| = � → ���� ∈ �(�, � = �).

��� � = |� − �| = ��, � = |� − �| = ��, � = |� − �| = ��.

(∗) ↔ � ��

�+

�� + � �

�+

�� + � �

�+

�� = �

↔ �(� + �) + �(� + �) + �(� + �) = �

↔ (� + � + �)(� + � + �) = �� + �� + ��

↔ � + � + � =�� + �� + ��

� + � + �↔ �� = �� ↔ � ≡ �

���� ���� �� �����������.

JP.369. Find �, �, � ∈ ��,�

�� such that:

���(��)

����+

���(��)

����+

���(��)

����+

��

�= �

Proposed by Daniel Sitaru-Romania

Solution 1 by proposer

���(��)

����+

���(��)

����+

���(��)

����+

��

�=

��

�+ �

���(��)

�������

=

= � ��

�+ �

���(��)

��������

� = � � ����(��)

�����+

��

���

=

= � � �������� − ������� + �����

�����+

��

���

=

= � � �������� − ������� + �

�+

��

���

= � �������� − ������� + �� + �

�����

=

=�

���(������� − ������� + ��)

���

=�

��(������ − �)�

���

= �

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→ ������� − � = ������� − � = �

������ − � = �

⎩⎪⎨

⎪⎧����� =

����� =�

����� =�

⎩⎪⎪⎪⎨

⎪⎪⎪⎧

���� = ±��

���� = ±��

���� = ±��

Therefore,

�, �, � ∈ ������ ���

�� ; � − ����� ��

���

Solution 2 by Daniel Văcaru-Romania

���(��)

����+

���(��)

����+

���(��)

����+

��

�; (∗)

Using ��� �� = � ���� � − �, ��� �� = � ���� � − � ��� � and

���(� + �) = ��� � ��� � − ��� � ��� �, we obtain

��� �� = �� ���� � − �� ���� � + � ��� � and

��� ��

��� �= �� ���� � − �� ���� � + �; (�)

We write (∗) as ∑ ��� ���� � − �� ���� � +��

�� = � ⇔

∑ �� ���� � −�

��

= �. For �, �, � ∈ ��,�

�� we obtain: ���� � =

→ ��� � =√�

�√�→ � = � = � = ����� �

√�

�√�� ∈ ��,

��

JP.370. In ���� the following relationship holds:

√�� + √�� + √�� + ��� + ��

�+ �

�� + ��

�+ �

�� + ��

�≤ �√��

Proposed by Daniel Sitaru-Romania

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Solution 1 by proposer

First we prove that ∀�, � > 0 then:

√�� + ��� + ��

�≤ � + �; (�)

Denote �� =�

�→ � = ���

���� ⋅ � + ����� + ��

�≤ ��� + �, � + �

�� + �

�≤ �� + �

��� + �

�≤ (�� − � + �)� ,

�� + �

�≤ (�� − � + �)�

�� + � ≤ �(�� + �� + � − ��� + ��� − ��)

��� + ��� + � − ��� + ��� − �� − �� − � ≥ �

�� − ��� + ��� − �� + � ≥ �, (� − �)� ≥ �

By (�) it follows that:

� �√�� + ��� + ��

��

���

≤ �(� + �)

���

= �(� + � + �) =

= � ⋅ �� = �� ≤⏞����������

� ⋅�√��

�= �√��

Solution 2 by Daniel Văcaru-Romania

�� ���: � ����

���

≤ �� ���

���

⋅ �� ���

���

, �� ����:

√�� + √�� + √�� + ��� + ��

�+ �

�� + ��

�+ �

�� + ��

�=

= � ∙ √�� + � ∙ √�� + � ∙ √�� + � ∙ ��� + ��

�+ � ∙ �

�� + ��

�+ � ∙ �

�� + ��

�≤⏞

���

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≤ �� ��� + �� + �� +�� + ��

�+

�� + ��

�+

�� + ��

�� =

= ��(�� + �� + �� + �� + �� + ��) ≤ √�� ⋅ ��� + �� + ��

But:

�� + �� + �� ≤ ��� → √�� + �� + �� ≤ ��. We have:

√�� + √�� + √�� + ��� + ��

�+ �

�� + ��

�+ �

�� + ��

�≤ �√���� + �� + �� = �√��

JP.371 Solve for real positive numbers the equation

����� + ����� + ����� = �����

Proposed by D.M. Bătineţu-Giurgiu ,Neculai Stanciu-Romania

Solution 1 by proposers

It is well known that ����� = ����� for any real positive number �.

So, the given equation becomes ����� + ����� + ����� = �����

, then we denote ���� = �, and the equation becomes

�� + �� + �� = �� ⇔ ��

��

+ ��

��

+ ��

��

= �,

and since LHS is decreasing, it results that we have unique solution.

We note that � = �.

Therefore ���� = �and the solution is � = ��.

Solution 2 by Ravi Prakash-New Delhi-India

Let � = ���� � → ��� �� = ��� � ⋅ ��� � = �������� �� → �� = ���� � → ���� � = ���� �

Similarly for other three expressions. The equation becomes:

�� + �� + �� = ��, ����� � = ��� � , ( � > �); (�)

Now, we can rewrite as:

��

��

+ ��

��

+ ��

��

= �; (�)

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Let �(�) = ��

��

+ ��

��

+ ��

��

, � ∈ ℝ

��(�) = − ��

��

��� � − ��

��

��� ��

�� − �

��

��� ��

�� < �, ∀� ∈ ℝ →

� −is strictly decreasing on ℝ. Thus, equation (�) and (�) have at most one solution.

So, � = � is the only solution of (�), thus

��� � = � → � = ��,where � is base of the logarithm in the question.

Solution 3 by Daniel Văcaru-Romania

(∗): ����� + ����� + ����� = �����

We have ���� � = ������ ����� �

= ������ ����� �

= ���� �; (�)

Taking account of (�), we have:

(∗) ⟺ ���� � + ���� � + ���� � = ���� �; (�)

Setting ��� � =⏞���

�, we obtain equation �� + �� + �� = �� ⟺ ��

��

�+ �

��

�+ �

��

�= �

But: �: ℝ → ℝ�∗ , �(�) = �

��

�+ �

��

�+ �

��

� is decreasing. Then equation �(�) = � has an unique

solution, � = � ⟺ ��� � = � ⟺ � = ����.

JP.372. If in ����: �� + �� = ��� then:

���� + ��� ⋅ �

��

���� ≤

√�

�(�� + �� + ��)

Proposed by Daniel Sitaru – Romania

Solution by proposer

����� = �� ⋅

�(�� + ��) − ��

�=

����� + ����� − ��

�=

=������������������

�=

����������������

�=

�����

� (1)

����� = �� ⋅

�(�� + ��) − ��

�=

����� + ����� − ��

�=

=������������������

�=

����������������

�=

�����

� (2)

By (1); (2) ⇒ ����� = ����

� ⇒ ��� = ��� (3)

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��

���

=��

��(�� + ��) − ��

� �� =

����

�(�� + ��)� + �� − ���(�� + ��)=

=� ⋅ (���)�

�(�� + ��)� + ��� + ��

� ��

− (�� + ��) ⋅ �(�� + ��)

=

=��������

����������������

=�

���

=��

� (4)

����

�����

�=

����

�(�� + ��) − ��

� ⋅�(�� + ��) − ��

=

=������

�� ��� +�� + ��

� � − ��� �� ��� +�� + ��

� � − ���=

=������

(��� + �� + �� − ��)(��� + �� + �� − ��)=

=������

����� =��

� (5)

By (4); (5) ⇒��

��� =

����

�����

� ⇒��

��� =

��

����⇒

⇒ � = �����

���� (6)

��� + ��� + ��� ≤���

�(�� + �� + ��)���� + ��

� + ����

By (3); (6):

��� + ��� + �� ⋅ �����

����≤ �(�� + �� + ��) ⋅

�(�� + �� + ��)

���� + ����

��

����≤

√�

�(�� + �� + ��)

Equality holds for � = � = �.

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JP.373 If �, �, � < �; � + � + � = �; �� – Fibonacci numbers; �� – Lucas

numbers; �� – Pell numbers; � ∈ ℕ; � ≥ � then:

��(�� − ��)(�� − ��)

����+

��(�� − ��)(�� − ��)

����+

��(�� − ��)(�� − ��)

����≥ �

Proposed by Daniel Sitaru – Romania

Solution by proposer:

Lemma: If �, �, � > � different in pairs then:

��

(���)(���)+

��

(���)(���)+

��

(���)(���)= � (1)

Proof:

��

(� − �)(� − �)+

��

(� − �)(� − �)+

��

(� − �)(� − �)=

=−��(� − �) − ��(� − �) − ��(� − �)

(� − �)(� − �)(� − �)=

=−��� + ��� − ��� + ��� − ��� + ���

(�� − �� − �� + ��)(� − �)=

=��� + ��� + ��� − ��� − ��� − ���

��� + ��� + ��� − ��� − ��� − ���= �

We take in (1): � = ��; � = ��; � = ��

����

(�����)(�����)+

����

(�����)(�����)+

����

(�����)(�����)= � (2)

��(�� − ��)(�� − ��)

����+

��(�� − ��)(�� − ��)

����+

��(�� − ��)(�� − ��)

����=

=��

����

(�� − ��)(�� − ��)

+��

����

(�� − ��)(�� − ��)

+��

����

(�� − ��)(�� − ��)

≥��������� (� + � + �)�

����

(�� − ��)(�� − ��)+

����

(�� − ��)(�� − ��)+

����

(�� − ��)(�� − ��)

=(�)

=(� + � + �)�

�= �� = �

Equality holds for � = � = �.

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JP.374 Solve for complex numbers:

���� − ����� + ����� − ����� + ���� − ��� + � = �

Proposed by Daniel Sitaru – Romania

Solution by proposer

���� − ����� + ����� − ����� + ���� − ��� + � = �

(���� − ��� + ��) + ����� − ������ + (����� − ���) +

+(�� − �����) + ���� − ��� + � = �

����� − ����� + ����� − ����� + ���� − ��� + �� +

+�� + ��� − ��� + ���� − ���� = �

���

� (��)� − ���

� (��)� + ���

� (��)� − ���

� (��)� + ���

� (��)� −

− ���

� �� + ���

� + �� + ��(� − �� + ���� − ���) = �

(� − ��)� + �� + ��(� − ��)� = �

�� − ��

��

+ �� − ��

��

+ � = �

��

�− ��

+ ��

�− ��

+ � = �

�− � = � ⇒ �� + �� + � = �

We denote �� = �; �� + � + � = �, ��,� = −�

�± �

√�

�� = �����

�+ � ���

��

�; �� = ���

��

�+ � ���

��

�� = �����

�+ � ���

��

�� = ���

��� + ���

�+ � ���

��� + ���

�; � ∈ {�, �, �}

�� = �����

�+ � ���

��

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��� = ���

��� + ���

�+ � ���

��� + ���

�; � ∈ {�, �, �}

Solutions:

�� =�

�� + �; ��

� =�

��� + �

; � ∈ {�, �, �}

JP. 375 Let �, �, � be positive real numbers such that � + � + � = �.

Prove that:

�(�� + �� + ��) − �(�� + �� + ��) ≥ ��

Proposed by George Apostolopoulos – Messolonghi – Greece

Solution 1 by proposer

It is well-known the identity

(−� + � + �)� + (� − � + �)� + (� + � − �)� = (� + � + �)� − �����

So (� − ��)� + (� − ��)� + (� − ��)� = �� − �����

Using the AM-GM inequality, we have:

�����

�≥ √���

� so ��� ≤ �

Now,

(�� − ��� + ���� − ���) + (�� − ��� + ���� − ���) +

+(�� − ��� + ���� − ���) ≥ �� − ��

Or

−��(� + � + �) + ��(�� + �� + ��) − �(�� + �� + ��) ≥ � − � ⋅ �� so

��(�� + �� + ��) − �(�� + �� + ��) ≥ � − � ⋅ �� + � ⋅ �� = ��

Namely �(�� + �� + ��) − �(�� + �� + ��) ≥ ��

Equality holds when � = � = � = �.

Solution 2 by George Florin Șerban-Romania

� = � �

���

= �, � ��

���

= �� �

���

− � � ��

���

= � − ��; � = � ��

���

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� ��

���

= �� �

���

− � �(� + �)

���

= �(�� − ��) + �� =; (� = ���)

= �(� − ��) + �� = �� − �� + ��

� � ��

���

− � � ��

���

= �(� − ��) − �(�� − �� + ��) = �� − �� ≥⏞(∗)

��

(∗) ⇔ �� ≤ � → � ≤ �; (∗∗)

��� ≤ �� + � + �

��

= ��

��

= � → � ≤ � → (∗∗) → (∗) ����.

Solution 3 by Samir Cabiyev-Azerbaijan

(� + � + �)(�� + �� + ��) = �� + �� + �� + ��� + ��� + ��� + ��� +

��� + ��� + (��� + ��� + ���) − ���� =

= �� + �� + �� + ��(� + � + �) + ��(� + � + �) + ��(� + � + �) − ���� =

= �� + �� + �� + (� + � + �)(�� + �� + ��) − ����

�� + �� + �&� = (� + � + �)� − �(�� + �� + ��)

Else, �(�� + �� + ��) − �(�� + �� + ��) =

= �(� + � + �)� − ��(�� + �� + ��) −

−� �(� + � + �)(�� + �� + ��) − (� + � + �)(�� + �� + ��) − ����� =

= � ⋅ �� − ��(�� + �� + ��) − �(�� + �� + �� − (�� + �� + ��) − ���) =

= �� − ��(�� + �� + ��) − �((� + � + �)� − �(�� + �� + ��) − ���) =

= �� − �� − ��(�� + �� + ��) + ��(�� + �� + ��) − ���� =

= �� − ����; � + � + � = �

Equality holds for � = � = � = �

Solution 4 by Nikos Ntorvas-Greece

�� + �� + �� = (� + � + �)� − �[(� + � + �)(�� + �� + ��) − ���]

�� + �� + �� = (� + � + �)� − �(�� + �� + ��)

We denote: � = � + � + �, � = �� + �� + ��, � = ���

��� = �(�� − ��) − �[�� − �(�� − �)] ≥ ��

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��� − ��� − ��� + ��� − �� ≥ ��

� ⋅ �� − ��� − � ⋅ �� + ��� − �� ≥ ��

−�� + �� ≥ �� ⇔ � − � ≤ � which is true from:

� + � + � ≥ �√����

⇔ � ≥ �√����

⇔ � ≥ √����

⇔ � ≥ ��� = �.

Equality holds for � = � = � = �

Solution 5 by Angel Plaza-Spain

By changing variables by � = ��, � = ��, � = ��, the problem becomes to proving that

��(�� + �� + ��) − ��(�� + �� + ��) ≥ ��, where �, �, � are positive numbers with

� + � + � = �. The inequality becomes homogeneous by multiplying by the appropriate

power of (� + � + �):

��(�� + �� + ��)(� + � + �) − ��(�� + �� + ��) ≥ ��(� + � + �)�, which is equivalent

to ��� + ��� + ��� + ����(� + �) + ����(� + �) + ����(� + �) ≥ ������ which

follows by the weighted arithmetic mean geometric mean inequality.

SP.361 Let (��)���, �� = � such that

��[�(���� − �� − �) − ��] + ��� = �[�(�� + ��) − ����]

Find:

����→�

�� +�

���

���

�� + �

���

��

��

Proposed by Florică Anastase-Romania

Solution by proposer

��[�(���� − �� − �) − ��] + ��� = �[�(�� + ��) − ����]

���(���� − �� − �) + �(���� − ���) + ��� = �� + ��� ⇔

�� + �

(� + �)(� + �)���� −

�� − �

�(� + �)�� = �

� ��� + �

(� + �)(� + �)���� −

�� − �

�(� + �)���

���

=�(� + �)

�⇔

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�� =�(� + �)(�� − � + �)

�(�� − �)

�� = ���

��

�� + �

���

�� − ���� = ���

��

�� + �

���

− �����

��

�� + �

���

���

=

=�

�� + �+ �

���

� − �����

�� + �

���

���

=�

�� + �+ �

�������

�� + �

���

���

= �����

����

�� + �

���

=

= �

�� ��

��

�� + �

���

=�

���

��

�� + ��

��

���

=�

��� �� −

�� + �� �

��

���

=�

��(�� − ��) ⇒

�� +�

��� �� = ���� +

��

��⇒ �� =

��

�� + ����� +

��

�� + �

�(�): �� ≥��

��, �(�): �� ≤

��

� − �, ∀� ≥ �

�(�): �� ≥��

�⇔ �

���

�� + �

���

≥ � (����)

If �(� − �), � ≥ � is true, thus

�� =��

�� + ����� +

��

�� + �>

��

�� + �∙

����

� − �=

��

�� + ��

� + �+ �� =

=��(����)

�(����)��>

��

�⇒ �(�) is true for all � ≥ �.

�(�): �� ≤��

�⇔ �

����

�� + �

���

≤��

� (����).

If �(� − �), ∀� ≥ � is true, thus

�� ≤��

�� + �∙

����

� − �+

��

�� + �=

��

�� + ��

� − �+ �� =

��

�� + �∙

�� − �

� − �=

=��(����)

(���)(����)����≤

��

���⇒ �(�) is true for all � ≥ �. So,

��

�≤ �

���

�� + �

���

≤��

� − �⇒

�≤

���

���

�� + �

���

≤�

� − �

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�� +�

��

��

��

≤ �� +�

���

���

�� + �

���

��

��

≤ �� +�

� − ��

��

��

����→�

�� +�

��

��

��

= ����→�

�� +�

��

�(���)��������

���(����)= ���

�→���� +

��

(���)(������)

���(����)= √�

����→�

�� +�

� − ��

��

��

= ����→�

�� +�

� − ��

�(���)��������

���(����)=

= ����→�

��� +�

� − ��

���

�(���)(������)

���(����)(���)

= √��

⇒ � = ����→�

�� +�

���

���

�� + �

���

��

��

= √��

SP.362 Let ��, ��, �� be the lengths of the medians of a triangle ∆���.

Prove

�√�

��≤

����

��+

����

��+

����

��≤

√��

���

Proposed by George Apostolopoulos-Messolonghi-Greece

Solution 1 by proposer

Let �, �, � be the lengths of the sides ��, ��, ��,

respectively. We have:

�� + �� = ���� +���

�= ���� + ����

= �(��� + �� ⋅ ��) =

= �(��� + �� ⋅ ���) = ���(�� + ���)

= ��� ⋅ ���

So, �� + �� = ��� ⋅ ���; ��� ≤ ��. Namely,

�� + �� ≤ ���� or

���� ≥ �� + �� ≥ ���; (�� �� − ��). We know

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that: � =�

��� ⋅ ����. So,

���� ≥��

���� or

������≤

�. Also, we have: � = �� and � ≥ �√��. So,

������≤

��≤

√��

��� or ����

��≤

√��

��� , and similarly: ����

��≤

√��

��� ,����

��≤

√��

���

Now, using AM-GM inequality, we have:

������+

������+

������≥

�������� ⋅ √������������

� ≥

≥�

�� + �� + ��

� ⋅���� + ���� + ����

=��

(�� + �� + ��)(���� + ���� + ����)

We know that ��� + ��

� + ��� =

�(�� + �� + ��) ≤

�⋅ ��� =

��

���, so

(�� + �� + ��)� ≤ �(��� + ��

� + ���), namely

(�� + �� + ��)� ≤��

��� or �� + �� + �� ≤

��. Also, it is well-known that:

���� + ���� + ���� ≤�√�

Hence, ����

��+

����

��+

����

��≥

��

��⋅

�√�

=�√�

��

Therefore,

�√�

��≤

����

��+

����

��+

����

��≤

√��

���

Solution 2 by Aggeliki Papaspyropoulos-Greece

�√�

��≤

����

��+

����

��+

����

��≤

√��

���; (�)

��� � =�

��� �; ��� � =

��� �; ��� � =

��� �

(�) →�√�

��≤ �

�� ⋅ ��� ����

≤√��

���; (�)

�√�

��≤ �

��

� ⋅ �����

≤√��

���; (�)

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(�): �√�

��≤ �

��

� ⋅ �����

⇔ ��

� ⋅ �����

≥�√�

���; (�)

� ⋅ �� ≤�

��⋅ ���; �∵ �� ≤

����(����������)�

→ � ⋅ �� ≤� ⋅ �

�, ( � = ���� �� ����) →

� ⋅ ��≥

� ⋅ �=

� ⋅ ��=

��

��

� ⋅ �����

≥�

��.

So it is enough to prove that: �

��≥

�√�

��� , ���� (�) → �� ≥ ��√�; (�)

�≥

�√�; (����������)

��, � ≥��

�√�→ �� ≥

���

�√�=

��

√�= �√�� → �� ≥ ��√� → (�) �� ����.

(��): ���

� ⋅ �����

≤√��

���⇔ �

� ⋅ �����

≤√�

���

∵ �� ≥�� + ��

��, �� ����:

� ⋅ �� ≥�(�� + ��)

��→

� ⋅ ��≤

��

�(�� + ��)≤

��

����=

��

���

→ ��

� ⋅ �����

≤��

���=

��

�� ⋅ ��=

� ⋅ ��

So, we have to prove: �

���≤

√�

��� ⇔√�

�≤

��⇔ � ≤

�√� (����������) → (��) is true.

SP.363 Triangle ��� has |��| = �, |��| = �, |��| = �,inradius � and

circumradius �. Equilateral triangles ����, ���� and ���� with centroids

�, � and � respectively, are drawn externally to triangle ���. Prove that:

�√� ≤ [���] + [���] + [���] ≤�√�

���

Proposed by George Apostolopoulos-Messolonghi-Greece

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Solution 1 by proposer

We have:

�� = �� =�

�⋅

�√�

�=

�√�

�,

�� = �� =�√�

� and �� = �� =

�√�

∠��� = ��° + � + ��° = ��° + �

∠��� = ��° + �, ∠��� = ��° + �

Also, we have:

[���] =�

��� ⋅ �� ⋅ ���(���) =

=�

�⋅

�√�

�⋅

�√�

�⋅ ���(��° + �) =

=��

��

√�

����� +

������ =

��

��

√�

�⋅

�� + �� − ��

���+

�⋅

��� =

=√�

��(�� + ��� − ��) +

���

� ⋅ ��=

√�

��(�� + �� − ��) +

[���]

So, [���] =√�

��(�� + �� − ��) +

[���]

�. Similarly, we have:

[���] =√�

��(�� + �� − ��) +

[���]

� and [���] =

√�

��(�� + �� − ��) +

[���]

�. So,

[���] + [���] + [���] =√�

��(�� + �� + ��) +

[���]

We know that �� + �� + �� ≤ ���, then � = �� − �� = (� − �) + (� − �) + (� − �)

By AM-GM we have: � ≥ ��(� − �)(� − �)(� − �)�⇔ �� ≥ ��(� − �)(� − �)(� − �)

⇔ �� ≥ ���(� − �)(� − �)(� − �). We know that: [���] = ��(� − �)(� − �)(� − �)

�� ≥ ��[���]� ⇔ �� ≥ �√�[���] ⇔ [���] ≤��

�√�. Also, we know that:

� =�√�

��, so [���] ≤

�√�

���. Now, we have:

[��� + [���] + [���] =√�

��(�� + �� + ��) +

[���]

�≤

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≤√�

��⋅ ��� +

�√�

�=

�√�

���

Also, we have: [��� + [���] + [���] =√�

��(�� + �� + ��) +

[���]

�≥

≥√�

��⋅ ���� +

��

�≥

√�

��⋅ ���� +

���√���

�=

�√�

��� +

�√�

��� = �√���

Therefore,

�√� ≤ [���] + [���] + [���] ≤�√�

���

Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco

�� ���� ∶ �� =

���

��� ���=

� �����

=√�

��, ��������� ∶ �� =

√�

��, ��� µ(���) = � +

�.

→ [���] =�

�. ��. ��. ��� ��� =

�.√�

��.

√�

��. ��� �� +

��

=�

��� �

���� � +

√�

���� �� =

=�

��.���

��+

√�

����.

�� + �� − ��

���=

�� +

√�

��(�� + �� − ��). (∴ � = [���])

→ [���] =�

�� +

√�

��(�� + �� − ��) (��� �������)

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→ �[���] = � ��

�� +

√�

��(�� + �� − ��)� =

�� +

√�

��� ��.

���, �[���] ≥⏞������� �����������

�� +

√�

��. �√�� = � = �� ≥⏞

����������

�√��� (�).

��� ∶ �[���] ≤⏞�������

��� +

√�

��. ��� ≤⏞

���������������

�.�√�

��.

�� +

�√�

���

=�√�

��� (�).

(�), (�) → �√��� ≤ �[���] ≤�√�

���.

SP.364 Let ��� be a non-right triangle with circumradius �. Squares with

sides ��, ��, �� and centroids �, �, � respectively, are drawn externally to

triangle ���. Let �, �, � be the distance from the vertices �, �, � to the

segments �������, ������, �������, respectively. Prove that:

�����

��

+ �����

��

+ �����

��

≥� ⋅ �����°

���

Proposed by George Apostolopoulos-Messolonghi-Greece

Solution 1 by proposer

Let �, �, � be the lengths of the sides

��, ��, �� of ∆���, respectively.

We have �� =�√�

�, �� =

�√�

� and using the

law of the cosines in ∆���, we have:

��� = ��� + ��� − ���

⋅ �����(��° + �)

=�� + ��

�+ �� ⋅

��=

�� + ��

�+ ��,

Where � = ����(����) =���

��.

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So, ��� =�����

�+ ��, ��� =

�����

�+ ��, ��� =

�����

�+ ��

Now, we have ����(����) =�

��� ⋅ � =

�������

�+ �� =

��� ⋅ �� ⋅ ��� (���)

But �������

�+ �� =

�√�

�⋅

�√�

�⋅ ����, so �������

�+ �� =

��

�⋅

��������

���

� ⋅ ��� + ��

�+ �� =

�� + �� − ��

�⇔ � =

�� + �� − ��

���� + ��

� + ��

≤�� + �� − ��

�√�� + ��

We have: � =�

��� ⋅ ���� ⇔ �� =

��

���� or � ≤

��������

����

�������

⇔ ���������

��

≥ ��� ��

����+ ��

Similarly, ���������

��

≥ ��� ��

����+ �� and �

��������

��

≥ ��� ��

����+ ��.

Adding up these inequalities, we have:

��� + �� − ��

��

+ ��� + �� − ��

��

+ ��� + �� − ��

��

≥ ��� ��

����+

����+

����+ ��

We know that: �

����+

����+

����≥ �√�, in any ����.So,

��� + �� − ��

��

+ ��� + �� − ��

��

+ ��� + �� − ��

��

≥ ����√� + ��

From the law of cosines in ����, we have:

�� = �� + �� − ��� ⋅ ���� ⇔ �� + �� − �� = ��� ⋅ ����

�� + �� − �� = � ⋅��

����⋅ ���� = �� ⋅ ����and similarly,

�� + �� − �� = �� ⋅ ����, �� + �� − �� = ������. So,

�������

��

+ �������

��

+ �������

��

≥ �����√� + �� ⇔

�����

��

+ �����

��

+ �����

��

≥���√� + ��

Let �� = � + � + � is the perimeter of the ����, then

Page 34: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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� = �� − �� = � + � + � − � − � − � = (� − �)(� − �)(� − �)

By AM-GM: � ≥ ��(� − �)(� − �)(� − �)�⇔ �� ≥ ��(� − �)(� − �)(� − �)

⇔ �� ≥ ���(� − �)(� − �)(� − �). We know that � = ��(� − �)(� − �)(� − �), so

�� ≥ ����or� ≤��

�� and � ≤

�√�

��, namely � ≤

�√�

���. Now, we have:

�����

��

+ �����

��

+ �����

��

≥���√� + ��

�≥

�√� + �

�√�� ��

=���� + √��

���=

��� + √��

���

Also, we know that �����° = � + √�, so

�����

��

+ �����

��

+ �����

��

≥� ⋅ �����°

���

Equality holds if and only if triangle is equilateral.

Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco

�� ���� ∶ �� =√�

��, �� =

√�

�� ��� µ(���) = � +

�.

→ [���] =�

�. ��. ��. ���(���) =

���. ��� �� +

�� =

���. ��� �.

��� ∶ ��� = ��� + ��� − �. ��. ��. ���(���) =�

�(�� + ��) − ��. ��� �� +

�� =

=�

�(�� + ��) + ��. ��� � =

�(�� + ��) + ��.

��=

�(�� + ��) + ��. (∴ � = [���])

Page 35: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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→ �� = ��. [���]

���

=����. ���� �

�(�� + �� + ��)→ �

��� �

��

=�(�� + �� + ��)

����. ���� �

=���(�� + �� + ��)

(���)�→ �

��� �

��

=�� + �� + ��

��� (��� �������)

→ � ���� �

��

= ��� + �� + ��

���=

���� �� + ��� ≥⏞

������� �����������

�√�� + ��

��=

=�√� + �

�� ≥⏞

���������������

�√� + �

�√��� .

��

= ��� + √��

���

�� ���� ∶ ���� ��° =� − ���(�. ��)

� + ���(�. ��)=

� − ���(�� + ��)

� + ���(�� + ��)=

� + ��� ��°

� − ��� ��°=

� + √�

� − √�

= �� + √���

→ ��� ��° = � + √� → � ���� �

��

≥�. ��� ��°

���.

Solution 3 by Aggeliki Papaspyropoulou-Greece

Working in triangle ���, ∠��� = ��° + �

�� =�√�

�, �� =

�√�

��� = ��√�

��

+ ��√�

��

− � ⋅�√�

�⋅

�√�

�⋅ ��� ����

��� =�

�(�� + ��) − �� ⋅ (− ��� �)

→ ��� =�

�(�� + ��) + �� ⋅ ��� �

Similarly,

��� =�

�(�� + ��) + �� ⋅ ��� �

��� =�

�(�� + ��) + �� ⋅ ��� �

� ⋅ �� =��

���� �

�+ �� =

��

���� �

→ � =�� ⋅ ��� �

���→

�=

���

�� ⋅ ��� �

Page 36: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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��,��� �

�=

��� �

��� �⋅

���

�� ⋅ ��� �=

���

�� ⋅ ��� �=

���

��=

��

Because �� ⋅ ��� � = ��, � = [���] −area of triangle ���.

Similarly, ��� �

�=

��

� and

��� �

�=

��

� ���� �

��

���

=��� + ��� + ���

��; (�� = �� ⋅ ��� � = �� ⋅ ��� �)

→ � ���� �

��

���

=�

���

�� + ��

�+ �� +

�� + ��

�+ �� +

�� + ��

�+ ���

We have to prove:

�� + �� + �� + �� ≥��� ⋅ ��� ��°

���; (�)

��� ��° = � + √�; (�), ��� ≥�� + �� + ��

�; (�)(�������)

So, by (�), (�) it is enough to prove that:

(�� + �� + ��)�

�+ �� ⋅

�� + �� + ��

�≥ ��� ⋅ �� + √��; (�)

⟺(�� + �� + ��)�

��+

�(�� + �� + ��)

�≥ ���� + √��; (�)

Let � =��������

�, then (�) ⟺ �� + �� − ���� + √�� ≥ �; (�)

⟺ �� − �√���� + � + �√�� ≥ �; (�)

�� + �� + �� ≥ �√�� → � ≥ �√� → (�) is true.

SP.365. Let �, �: [�, �] → ℝ, such that �(�) = −�(�) + ����(�). If � −is

Riemann integrable, prove that � −is Riemann integrable.

Proposed by Cristian Miu-Romania

Solution by proposer

Let us prove first two things. The first things is that the function � → −� + ���� is

injective. The second thing is that if (��)� −is a sequence and – �� + ���� is convergent,

then (��)� −is convergent.

Page 37: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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To ptove that � → −� + ���� is injective we shall prove that � → � + ���� is injective.

From �� + ����� = �� + ����� we have:

������ − ��

����

�� + ��

�� = �

�� − ��

��

From here we obtain �� = ��, because |����| ≤ |�| with equality if and only if � = �.

So � → � + ���� is injective and because � →�

�− � is also injective, we obtain that

� → −� + ���� is injective.

Now if – �� + ����� is convergent, we obtain that (��)� −is bounded, so there exists

����→�

����� and ����→�

����� and they are real numbers.

Let � = ����→�

����� and � = ����→�

�����. Then exists ��� and ���

such as

����→�

���= � and ���

�→����

= �; ����→�

(−�� + �����) = �, so

����→�

�−���+ ������

� = � = ����→�

�−���+ ������

�. So – � + ���� = −� + ����.

But the function � → −� + ���� is injective. We obtain that � = � which prove that

(��)� −is convergent. Now, we shall use Lebesque’s theorem for Riemann integrability.

Because – �(�) + ����(�) is Riemann integrable, we obtain that � −is bounded.

Let �� −be a continuity point of �. We shall prove that �� is also a continuity point for �.

Let (��)� −be sequences such as ����→�

�� = ��. Then ����→�

�−�(��) + ����(��)� = �(��),

so we obtain that �(��) −is convergent.

So, ����→�

�(��) = �. W obtain that – � + ���� = −�(��) + ����(��) and because

� → −� + ���� is injective we obtain that � = �(��).

According to Lebesque theorem the function � −is Riemann integrable.

SP.366 Let ��, ��, �� be the medians, ��, ��, �� the exradii, � inradius and �

circumradius of a triangle ���. Prove that:

��

���

��

≤��

��� + ��

�+

���

��� + ��

�+

���

��� + ��

�≤ � �

���

−�

��

���

Proposed by George Apostolopoulos-Messolonghi-Greece

Page 38: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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Solution 1 by proposer

First we will prove that: �� ≤��

��. We have: �� = � ⋅ ���

�, where � −is the

semiperimeterof ����. Let �� = �, �� = �, �� = � and � − the area of ����. We

have:

��

��=

(������)�

� ��� �������

=�������

��=

�������

(������)(������)=

����

���������=

=����

�� ���

��

���(� − �) − ���(� + �)=

������ ���

��

���(� − �) + ����≥

������ ���

��

� + ����=

=����

�� ���

��

� + ������ �� − ��

=���

��

�����

= ����

So, ����

�≤

��

��=

��

���. Namely, �� ≤

��

��. Similarly, �� ≤

��

��, �� ≤

��

��. Now,

���

������

� ≤��

���������=

��

�����, because ��

� ≥ ����, ��� ≥ ����, ��

� ≥ ����. So,

���

��� + ��

�+

���

��� + ��

�+

���

��� + ��

�≤

��

�� + ��+

��

�� + ��+

��

�� + ��≤

��

���

� − � +�

� − �

+

��

���

� − � +�

� − �

+

��

���

� − � +�

� − �

=

=

��

��

� ��

� − � +�

� − ��

+

��

��

� ��

� − � +�

� − ��+

��

��

� ��

� − � +�

� − ��

=

=

��

��

� ⋅�

(� − �)(� − �)

+

��

��

� ⋅�

(� − �)(� − �)

+

��

��

� ⋅�

(� − �)(� − �)

=

=�(� − �)(� − �) + �(� − �)(� − �) + �(� − �)(� − �)

���=

=� �

� − � + � − �� �

+ � �� − � + � − �

� ��

+ � �� − � + � − �

� ��

���=

Page 39: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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=� ⋅

��

� + � ⋅��

� + � ⋅��

����

=�� + �� + ��

����

It is well-known that �� + �� + �� = ��(�� − ��� − ���) and

�� ≤ ��� + ��� + ���(���������). So,

���

��� + ��

�+

���

��� + ��

�+

���

��� + ��

�=

�� + �� + ��

����=

��(�� − ��� − ���)

��� ⋅ (��)=

=�� − ��� − ���

���≤

(��� + ��� + ���) − ��� − ���

���=

��� − ���

���=

= � ��

���

−�

��

���

Now, using Cauchy-Schwarz Inequality, we have:

���

��� + ��

�+

���

��� + ��

�+

���

��� + ��

�≥

(�� + �� + ��)�

�(��� + ��

� + ���)

We know that: �� + �� + �� = �� + �; �� + �� + �� ≤ ��� and

��� + ��

� + ��� =

�(�� + �� + ��) ≤

��

���

So,

���

��� + ��

�+

���

��� + ��

�+

���

��� + ��

�≥

(�� + �)�

� ⋅��� ��

≥� ⋅ ����

����=

��

���

��

Namely,

��

���

��

≤��

��� + ��

�+

���

��� + ��

�+

���

��� + ��

�≤ � �

���

−�

��

���

Solution 2 by Marian Ursărescu-Romania

For LHS, using Bergstrom’s inequality, we have:

���

��� + ��

����

≥(�� + �� + ��)�

����� + ��

� + ����

=(�� + �)�

� ⋅��

(�� + �� + ��)=

�(�� + �)�

�(�� + �� + ��)

We must show that:

�(�� + �)�

�(�� + �� + ��)≥

�⋅

���

��⇔ ��(�� + �)� ≥ ���(�� + �� + ��)

Page 40: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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But: �� + �� + �� ≤ ���; (�). From (�), (�) we must to prove:

��(�� + �)� ≥ ������ ⇔ (�� + �)� ≥ ���� ⇔ �� + � ≥ ��

�� ≥ �� ⇔ � ≥ �� (�����).

For RHS, using �� ≥ ��(� − �), we have:

��� + ��

� ≥ �(� − �) + �(� − �) = ��

We must show that:

�⋅ �

���

����

≤���

���−

��=

��� − ��

���; (�)

But: �� ≤��

��; (�). From (�), (�) we must show that:

� ⋅ ����⋅ � ��

���

≤��� − ��

���⇔

��⋅ � ��

���

≤ ��� − ��; (�)

But: ∑�� = ��(�� − ��� − ���); (�)

From (�), (�) we must show:

��⋅ ��(�� − ��� − ���) ≤ ��� − ��� ⇔

�� − ��� − ��� ≤ ��� − ��� ⇔ �� ≤ ��� + ��� + ���(���������)

SP.367. Let ��, ��, �� be the medians, ��, ��, �� the exradii, � the inradius

and � the circumradius of a triangle ���. Prove that:

��

��<

�� + ��

����+

�� + ��

����+

�� + ��

����≤

��� �

���

− ��

Proposed by George Apostolopoulos-Messolonghi-Greece

Solution 1 by proposer

First we will prove that: �� ≤��

��. We have: �� = � ⋅ ���

�, where � −is the semiperimeter

of ����. Let �� = �, �� = �, �� = � and � − the area of ����. We have:

��

��=

(������)�

� ��� �������

=�������

��=

�������

(������)(������)=

����

���������=

Page 41: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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=����

�� ���

��

���(� − �) − ���(� + �)=

������ ���

��

���(� − �) + ����≥

������ ���

��

� + ����=

=����

�� ���

��

� + ������ �� − ��

=���

��

�����

= ����

So, ����

�≤

��

��=

��

���. Namely, �� ≤

��

��. Similarly, �� ≤

��

��, �� ≤

��

��.Also, we have:

��� =

�(�� + ��) − ��

�≥

(� + �)� − ��

�=

(� + � + �)(� + � − �)

�=

=��(�� − ��)

�= �(� − �) =

�(� − �)(� − �)(� − �)

(� − �)(� − �)=

��

(� − �)(� − �)=

=�

� − �⋅

� − �= ����

Now, we have:

�� + ��

����≤

�� + ��

����� ⋅ �����

=�� + ��

�������

��

�� +��

��

�� − �

� �� − �

⋅�

� − �

=

�� + ��� ����

(� − �)�(� − �)(� − �)

=(�� + ��)(� − �)�(� − �)(� − �)

����≤

(�� + ��)(� − �) ⋅� − � + � − �

�����

=

=�������(���)�

���� . Similarly: �����

����≤

�������(���)�

���� ;�����

����≤

�������(���)�

����

�� + ��

����+

�� + ��

����+

�� + ��

����≤

≤�

�����(�� + ��)(� − �)� + (�� + ��)(� − �)� + (�� + ��)(� − �)�� =

=�

����⋅

(�� + ��)(� + � − �) + (�� + ��)(−� + � + �) + (�� + ��)(� − � + �)�

�=

=�

���⋅

�(���� + ���� + ����) + ��� + ��� + ��� + ��� + ��� + ��� − �(���� + ���� + ����)

�=

=�

��������(� + � + �) +

��(�� + ��) + ��(�� + ��) + ��(�� + ��)

�− (���� + ���� + ����)� ≤

Page 42: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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≤�

����

⎜⎛

���(� + � + �) +�

� + ��

��

(�� + ��) + �� + �

��

(�� + ��) + �� + �

��

(�� + ��)

�− (���� + ���� + ����)

⎟⎞

≤�

����

⎜⎛

���(� + � + �) +

�(�� + ��)�

�+

�(�� + ��)�

�+

�(��� + ��)�

��

− (���� + ���� + ����)

⎟⎞

≤�

��������(� + � + �) +

�� + �� + �� − (���� + ���� + ����)

��

Now, we will prove that ���� + ���� + ���� ≤�����

��

We have: ���� + ���� + ���� = (���)� ��

�� +�

�� +�

���

Also, we have (� − �)� ≥ � ⇔ �� − (� − �)� ≤ �� ⇔�

�� ≤�

���(���)� =

=�

(�����)(�����)=

(�����)(�����)=

�(���)(���). Similarly

�� ≤�

�(���)(���)and

�� ≤�

�(���)(���). So,

��+

��+

��≤

��

(� − �)(� − �)+

(� − �)(� − �)+

(� − �)(� − �)� =

=�

�⋅

� − � + � − � + � − �

(� − �)(� − �)(� − �)=

�⋅

�(�� − ��)

�(� − �)(� − �)(� − �)=

�⋅

��

��=

�����

=�

�⋅

��

���� =�

���. So, ���� + ���� + ���� ≤ (���)� ⋅�

���. Also we know that:

��� = �� ⋅ � = ����; �� + �� + �� = �(���� + ���� + ����) − ����. So,

�� + ��

����+

�� + ��

����+

�� + ��

����≤

��������� +

�����

��− ���� =

=���� ⋅ ��

�� ⋅ ����+

��

���−

�=

��+

��

���−

�=

��� + �� − ���

���≤

���� ���

���−

�=

=�

��� �

���

− ��. For the left inequality, we have:

�� + ��

����+

�� + ��

����+

�� + ��

����≥

����� ��(�� + ��)(�� + ��)(�� + ��)�

��������� �

� ≥

Page 43: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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≥��(�� + ��)(�� + ��)(�� + ��)

��� + �� + ��

� ��

=���(�� + ��)(�� + ��)(�� + ��)�

(�� + �� + ��)�

We know that: (�� + �� + ��)� ≤ ����� + ��

� + ���� = � ⋅

�(�� + �� + ��) ≤

≤ � ⋅�

�⋅ ��� =

��

���and (�� + ��)(�� + ��)(�� + ��) = ����, ; � ≥ �√��. So,

�� + ��

����+

�� + ��

����+

�� + ��

����≥

��√�����

����

≥������√���

�⋅ ��

����

=

=�(� ⋅ � ⋅ �)��

���=

��

��

Namely,

��

��<

�� + ��

����+

�� + ��

����+

�� + ��

����≤

��� �

���

− ��

Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco

�� ���� ���� �� ≤���

�� (����������) ���

��=

��+

��=

�� + ��

���� (��� �������)

→ ��� + ��

����≥ �

��

��

��� + ��

����= �

��

��

������

������= �

��

��

.�

������� ����

= ���

��

.��

���. �� =

�≥⏞

����� ��

��

���, �� ���� ���� �� ≥ �� (��� �������).

→ ��� + ��

����≤ �

�� + ��

����= �

�����

������=

������� ���� =

��

���. ��

=�

��≤⏞?

��� �

���

− ��

↔ � ��

���

− � ��

��� − � ≥ � ↔ (� − �)(��� + ��� + �� + �) ≥ � �∴ � =

���

����� �� ���� ���� ����� �� =�

��≥ ��.

Page 44: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

���������,

��

��≤ �

�� + ��

����≤

��� �

���

− ��

SP.368 If � < � < � < �, then prove:

�(�� − �)

�√�� + ��< �

��

��(�� + ��)�

+√�

�<

√�� + ��+

� − �

�√�

Proposed by Florică Anastase-Romania

Solution by proposer

Let be the function:

�, �: [�, �] → �, �(�) =�

� ș� �(�) =

�(�� + ��)� , �: [�, �] → �,

�(�) = �(�) − �(�) decreasing.

From second M.V.T. ∃� ∈ (�, �) such that:

� �(�)�(�)�� = �(�) � �(�)��

� �(�)��(�) − �(�)��� = (�(�) − �(�)) � �(�)��

� �(�)�(�)��

=

= �(�) � �(�)�� + ��(�) − �(�)� � �(�)�� =

�(�) � �(�)�� + �(�) � �(�)��

�(�) = � �(�)�� = ���

�(�� + ��)�= (∗)

⎩⎪⎨

⎪⎧

�� + �� = ����

� =�

√�� − �

�� = −����

�(�� − �)�

Page 45: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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(∗) = − ��(�� − �)�

��∙

��

�(�� − �)��� = −� �

��

��=

�=

��

√�� + ��+ ∁

���

��(�� + ��)�= �(�)(�(�) − �(�)) + �(�)��(�) − �(�)�

=�

√�� + ��+

�(� − �)

�√�� + ��−

√�

� < � < � →�(� − �)

�√�� + ��>

�(� − �)

�√�� + ��

Hence,

���

��(�� + ��)�

=�

√�� + ��+

�(� − �)

�√�� + ��−

√�

�>

√�� + ��+

�(� − �)

�√�� + ��−

√�

�=

=�(�� − �)

�√�� + ��−

√�

� < � < � →�

�� + ��<

���→

�(� − �)

�√�� + ��<

�(� − �)

��√�=

� − �

�√�

���

��(�� + ��)�

=�

√�� + ��+

�(� − �)

�√�� + ��−

√�

�<

√�� + ��+

� − �

�√�−

√�

Therefore,

�(�� − �)

�√�� + ��< �

��

��(�� + ��)�

+√�

�<

√�� + ��+

� − �

�√�

SP.369 Let (��)���, �� = �, �� = �, ���� = ���� + ��, ∀� ∈ ℕ, be the Lucas’

sequences, and �, �, � ∈ ℝ�∗ such that ��� = �. Prove that:

��(��� + �����)�+

��(��� + �����)�+

��(��� + �����)�≥

����� , ∀� ∈ ℕ

D.M. Bătineţu – Giurgiu, Neculai Stanciu-Romania

Solution 1 by proposers

By Bergström’s inequality we have

Page 46: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� =�

��(��� + �����)�+

��(��� + �����)�+

��(��� + �����)�≥

≥�

��

��(��� + �����)+

��(��� + �����)+

��(��� + �����)�

, ∀� ∈ ℕ; (�)

Also we have

�� =�

��(��� + �����)+

��(��� + �����)+

��(��� + �����)=

=

���

�(��� + �����)+

���

�(��� + �����)+

���

�(��� + �����), ∀� ∈ ℕ

Where we apply again the Bergström’s inequality and than AM-GM inequality and we

deduce that

�� ≥�

�� +

��

+���

(�� + �� + ��)(�� + ����)=

(�� + �� + ��)�

(���)�(�� + �� + ��)(���)�����=

=�� + �� + ��

����≥⏞

����� � ⋅ �(���)��

����=

����, ∀� ∈ ℕ; (�)

From (1) and (2) follows that

�� ≥�

��

�����

=�

�����

, ∀ � ∈ ℕ

Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco

��

��(��� + �����)�= �

����

(���� + ������)�=⏞

�����

��

���

��� . �� +

��

. ������ ≥

≥⏞�ö���� �∑

���

� �∑ ��� . �� +

��

. ������� =

�∑���

� �∑���

(���� + ��)�

=�∑

���

������ ≥⏞

�����

≥��

������ =

�����.

���������,

Page 47: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

��

��(��� + �����)�≥

����� , ∀� ∈ �

SP.370 If ��� is a triangle with inradius � and circumradius �, then for any

point � in the plane of triangle, � ∉ {�, �, �}, holds the inequality

��

�� + ��+

��

�� + ��+

��

�� + ��≥

� + �

�≥

��

Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

Solution 1 by proposers

In the inequality of Nesbitt �

���+

���+

���≥

�, ∀�, �, � ∈ ℝ; (�) we put � = ��, � =

��, � = ��, and we deduce that:

� =��

�� + ��+

��

�� + ��+

��

�� + ��≥

�; (�)

By Euler’ inequality, we have in any triangle : � ≥ �� ⇔�

�≥

�; (�)

From (�), (�) it follows that:

� ≥ � +�

�≥ � +

�=

� + �

�≥

�� + �

�=

��

Solution 2 by Daniel Văcaru-Romania

We have:

��

�� + ��+

��

�� + ��+

��

�� + ��=

=���

�� ⋅ �� + �� ⋅ ��+

���

�� ⋅ �� + �� ⋅ ��+

���

�� ⋅ �� + �� ⋅ ��≥⏞

���������

≥(�� + �� + ��)�

�(�� ⋅ �� + �� ⋅ �� + �� ⋅ ��)=

= � +��� + ��� + ���

�(�� ⋅ �� + �� ⋅ �� + �� ⋅ ��)≥⏞

�.�.� �

�= � +

�; (�)

But we know Euler, namely � ≥ ��; (�) we get: �

�≥

�; (�). Then we have:

��

�� + ��+

��

�� + ��+

��

�� + ��≥⏞(�)

�= � +

�≥⏞(�)

� +�

�=

� + �

�≥⏞(�)

��

Page 48: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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SP.371. Let ���� be a tetrahedron, and let � be a point in space,

� ∉ {�, �, �}. Prove that:

��

�� + �� + ��+

��

�� + �� + ��+

��

�� + �� + ��+

��

�� + �� + ��≥

� + �

�≥

��

Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania

Solution 1 by proposers

If �, �, �, � ∈ ℝ�∗ , � = � + � + � + �, then

� = ��

� − ����

≥�

�; (�)

� = ��

� − ����

= �� − � + �

� − ����

= −� + � ⋅ ��

� − ����

By Bergstrom’ inequality, we get:

� ≥ −� + � ⋅��

∑(� − �)= −� +

��

�=

We put in (�): � = ��, � = ��, � = ��, � = �� and we obtain

���

�� + �� + �����

≥�

�= � +

�; (�)

By the inequality of Euler-Durrande for tetrahedron we have: � ≥ �� ⇔�

�≤

� and then

by (�) we deduce that:

���

�� + �� + �����

≥ � +�

�≥ � +

�=

� + �

�≥

��

Solution 2 by Daniel Văcaru-Romania

��

�� + �� + ��+

��

�� + �� + ��+

��

�� + �� + ��+

��

�� + �� + ��=

=���

�� ⋅ �� + �� ⋅ �� + �� ⋅ ��+

���

�� ⋅ �� + �� ⋅ �� + �� ⋅ ��+

+���

�� ⋅ �� + �� ⋅ �� + �� ⋅ ��+

���

�� ⋅ �� + �� ⋅ �� + �� ⋅ ��≥⏞

���������

Page 49: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

≥(�� + �� + �� + ��)�

�(�� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ ��)=

= � +��� + ��� + ��� + ���

�(�� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ ��); (�)

���: �(��� + ��� + ��� + ���) ≥

≥ �(�� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ ��)

Hence,

��� + ��� + ��� + ���

�(�� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ ��)≥

�; (�)

On the other hand, we have � ≥ ��; (�) ⇔�

�≥

�; (�). Then we have:

��

�� + �� + ��+

��

�� + �� + ��+

��

�� + �� + ��+

��

�� + �� + ��=⏞(�)

= � +��� + ��� + ��� + ���

�(�� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ �� + �� ⋅ ��)≥⏞(�)

≥⏞(�)

� +�

�≥⏞(�)

� +�

�=

� + �

�≥⏞(�)

��

SP.372 If �: ℝ�∗ → ℝ�

∗ with ����→�

�(�)

�= � ∈ ℝ�

∗ , (��)��� is an arithmetic

progression with ��, � ∈ ℝ�∗ and �, � ∈ ℝ satisfy � + � = �, then compute:

����→�

�(� + �)� ���(��)�(��) … �(��)�(����)�����

− �� ���(��)�(��) … �(��)���

Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania

Solution 1 by proposers

We denote �(��)�(��) … �(��) = �(��)!, ∀� ∈ ℕ∗ and

�� = (� + �)� �(�(����)!)����− �� ���(��)�

��

= �� ���(��)���

(�� − �) =

= ���� �(�(��)!)��⋅

����

�� ��⋅ �� ��

� = ���(��)!

��

⋅����

�� ��⋅ �� ��

� , ∀� ≥ �, where

�� = ����

��

���(����)!

���

��(��)!� �

, ∀� ≥ �. We have

Page 50: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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����→��(��)

�= ����→�

�(��)

��⋅

��

�= �� and ����→�

�(����)!

��(��)!= ����→�

�(����)

�= ��. Also

we have that

����→�

��(��)!�

�= ���

�→��

�(��)!

��

= ����→�

��(����)!

(� + �)���⋅

��

�(��)!� = ���

�→��

��⋅

�(����)

� + �� =

��

�,

where�� = �� +�

��

→ �. We get

����→�

�� = ����→�

�� + �

��

⋅ ����→�

���(����)!

���

��(��)!�

=

= � ⋅ ����→�

���(����)!

���

� + �⋅

��(��)!�

⋅� + �

��

= ���

�⋅

��⋅ ��

= �, so ����→�����

�� ��= �. Hence

����→�

��� = ���

�→���

� ⋅ ����→�

��(����)!

�(��)!⋅

��(����)!���

=

= �� ����→�

��(����)

� + �⋅

� + �

��(����)!���

=

= �� ��� ⋅�

���

= ���� = �. Then

����→�

�� = ����→�

���(��)!�

��

⋅ � ⋅ �� � = ����→�

���(��)!�

��

= ���

��

Solution 2 by Mikael Bernardo-Mozambique

� + � = � → � − � = −�

� = ����→�

�(� + �)� ⋅ ���(��)�(��) … �(��)�(����)�����

− �� ⋅ ���(��)�(��) … �(��)���

� =

= ����→∞

⎛(� + �)� ⋅ ���(��)�(��) … �(��)�(��+�)�

��+�

− �� ⋅ ���(��)�(��) … �(��)���

� + � − �

⎞ =⏞��−�

= ����→�

�� ⋅ ���(��)�(��) … �(��)���

�= ���

�→����� ⋅ ���(��)�(��) … �(��)�

��

=

Page 51: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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= ����→�

���(��)�(��) … �(��)���

��; (����� � − � = �)

= ����→�

� ��(��) ⋅ �(��) ⋅ … ⋅ �(��)

��

=⏞���

�����→�

�(����)

(� + �)���⋅ ���

=

= �����→�

�� + � ⋅ �

� + �⋅

��

= �� ⋅ �

��

SP.373 If �: ℝ�∗ → ℝ�

∗ is a function such that ����→��(�)

�= � ∈ ℝ�

∗ and

(��)��� is a positive sequence such that ����→�(���� − ��) = � ∈ ℝ�∗ , then

compute:

����→�

�(� + �)�

��(��)�(��) … �(��)�(����)���−

��

��(��)�(��) … �(��)��

Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania

Solution by proposers

Denoting �(��)! = �(��)�(��) … �(��), ∀� ∈ ℕ∗, �� = �� +�

��

and

�� =(� + �)�

��(����)!���

−��

��(��)!�

=��

��(��)!�

(�� − �) =�

��(��)!�

⋅�� − �

�� ��⋅ �� ��

� , ∀� ≥ �

where �� = ����

��

� ��(��)!�

��(����)!��� . We have

����→�

��

�= ���

�→�

���� − ��

(� + �) − �= �; ���

�→�

�(��)

�= ���

�→��

�(��)

��⋅

��

�� = ��

����→�

��(��)!�

= ����→�

���

�(��)!

= ����→�

�(� + �)���

�(����)!⋅

�(��)!

��� = ���

�→��� ⋅

� + �

�(����)=

= � ����→�

�� + �

����⋅

����

�(��)� = � ⋅

�⋅

�=

��

����→�

�� = ����→�

��� + �

��

⋅��(��)!�

�⋅

� + �

��(����)!���

⋅�

� + �� = � ⋅

��

�⋅

��⋅ � = �

Page 52: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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So, ����→�����

�� ��= �

����→�

��� = ���

�→����

� ⋅�(��)!

�(����)!⋅ ��(����)!

���� = �� ⋅

��⋅

��

�= �

Hence � = ����→��

��(��)!� ⋅

����

�� ��⋅ �� ��

� =�

��⋅ � ⋅ �� � =

��

SP.374 Let ��, ��, �� be the lengths of the medians of a triangle with

circumradius R and area �.Prove that:

���≤

��(�� + ���)+

��(�� + ���)+

��(�� + ���)≤

√�

��

Proposed by George Apostolopoulos – Messolonghi – Greece

Solution 1 by proposer

For the right inequality we have:

First will prove that �

��(������)≤

���⋅ �

��+

���

We have �

���⋅ �

��+

��� −

��(������)=

=�

���

��

��+

��� −

�� + ���� =

��⋅

(��� + ��)(�� + ���) − �����

�����(�� + ���)=

��⋅

�(�����)�

�����(������)≥ �. Similarly

��(������)≤

����

��+

��� and

��(�� + ���)≤

����

��+

���

Adding up these inequalities, we get

��(�� + ���)+

��(�� + ���)+

��(�� + ���)≤

��

����+

����+

�����

Now, will prove that: �

����+

����+

����≤

√�

Page 53: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

We know that in triangle ���, we have ��� =�

���, �� =

���, ��� =

��� and

area ����� =�

��. Also we know that

� + � + � = �� ≤ �√�� or � + � + � ≤ �√����

��. (��� = �� ⋅ �)

So for ����, we have

��� +

��� +

��� ≤ �√� ⋅

���⋅

���⋅

���

�⋅�

��

or �

����+

����+

����≤

√�

�. So,

��(�� + ���)+

��(�� + ���)+

��(�� + ���)≤

√�

��

Now, for the left inequality, we have by Cauchy – Schwarz inequality, �

��(�� + ���)+

��(�� + ���)+

��(�� + ���)≥

(� + � + �)�

�(���� + ���� + ����)

Also we have

���� + ���� + ���� ≤ ��� + ��

� + ��� =

�(�� + �� + ��)

We know that �� + �� + �� ≤ ��� in any triangle ���. So

���� + ���� + ���� ≤��

���. Namely

��(�� + ���)+

��(�� + ���)+

��(�� + ���)≥

� ⋅��� ��

=�

���

Equality holds if and only if the triangle ��� is equilateral.

Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco

��

��(�� + ���)≥⏞

��� �

∑ ��(�� + ���)=

∑ ����≥⏞

∑ ���∑ ��

∑ ���

=�

��

∑ ��≥⏞

������� �

���.

���, �� ���� ���� ��, ��, ����� �� ��� ����� − ������� �� ��������.

��� ��, ��, ���� ��� ����, ������������� ��� ����������� �� �������.

Page 54: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

��

��(�� + ���)= �

���� + �����≤⏞

���

��

��

����+

����� =

��

����

=∑ ��

�������=

=��′

�. ��′�′≤⏞

���������� �√��′

�. ��′�′=

√�

��′=

√�

��. �∴ �� =

���

���������,�

���≤ �

��(�� + ���)≤

√�

��

SP.375 If �, �, � ∈ (�, �), then in any ��� triangle with the area � the

following inequality holds:

���

(� + �)�(� − ��)+

���

(� + �)�(� − ��)+

���

(� + �)�(� − �)≥ �√���

Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu – Romania

Solution 1 by proposers

Let be �: (�, �) → ℝ, �(�) = �(� − ��) with ��(�) = � − �� − ��� =

= � − ���, ��(�) = � ⇒ � =�

√�, � �

√�� =

√��� −

�� =

�√� and

� �

√� �

��(�) + + + + + + � − − − − − − −

�(�) ↗

�√� ↘

⇒ �(�) ≤�

�√�, ∀�(�, �)

and then

����

(� + �)�(� − ��)���

= �����

(� + �)��(�)���

≥�√�

��

����

(� + �)�

���

≥��������� �√�

�⋅

���

���

� + ����

≥���������� √�

���√���

�=

Page 55: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

=�√� ⋅ ���

�= �√���

Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco

From AM-GM, we have:

�� +√�

��+

√�

��≥ � ⋅ ��� ∙

√�

��∙

√�

��

= �

→ � − �� ≤�√�

�� (��� �������)

����

(� + �)�(� − ��)���

≥�

�√��

����

(� + �)�

���

=�√�

�� �

���

� + ��

���

≥⏞���

√�

���

���

� + ����

We know that:

��

� + ����

∙ �� ≥ �√�� (����������)

Therefore,

����

(� + �)�(� − ��)���

≥√�

���√���

�= �√���

UP.361 Prove that:

�������

�√� − ��

�� = ����(√� − �) � ���(����)

��

��

Proposed by Florică Anastase-Romania

Solution 1 by proposer

∵ � �����

���

���

=√�

����

Let: ��, � = �, �, . . , �� the roots of the unity.

�� = �����

��+ ����

��

��, � = �, �, … , ��

Page 56: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

��� − � = �(� − ��)

��

���

=⏞

��,��±�������

(�� − �) �(� − ��)(� − �����)

���

���

= (�� − �) � ��� − �������

�+ ��

���

���

⇒ ����� + ����� + ⋯ + �� + � = � ��� − �������

�+ ��

���

���

������

� = � �� − ������

��

���

���

= � ��������

���

���

���

� = ��(���) ∙ �����

��∙ ����

��

��∙ … ∙ ����

(� − �)�

��

���� ∙ ����

��∙ ���

��

��∙ … ∙ ���

(� − �)�

��= √� ⇒ � ���

��

���

���

=√�

����

� ���(����)

��

�� =�

�� ���(����)

�� =�

�� ���(�����)

�� =

= ����→�

��� ��� ����

��

��

���

���

= ����→�

����� �� ���

��

���

���

� =

= ����→�

����� �

√�

����� =

�����→�

���√� − (� − �)����

�= −

�����; (�)

�(�) = ��������

�√� − ��

�� ⇒ ��(�) = ���

(� + ����)√� − ��

= ���

� + �������

��

=�

�� + ������� �

����

�� + ��� =

��� + ��⇒

�(�) =�

���� �� + �� + ��� + �

�������

�√� − ��

�� =�

������ + √��; (�)

From (�), (�) we get:

Page 57: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

�������

�√� − ��

�� = ����(√� − �) � ���(����)

��

��

Solution 2 by Timson Azeez Folorunsho-Lagos-Nigeria

� = �����√� − �� � ���(����)

��

�� =⏞

�→��

��

�����√� − �� � ��� ���� ��

�− ��� (−��)

��

=

= �����√� − �� � ���(����)

��

��

�� = �����√� − �� � ���(���� ⋅ ����)

��

�� = �����√� − �� � ��� ������

��

��

�� =

= �����√� − ��(� ���(�����)

��

�� − � ���(�)

��

�� =⏞����

= �����√� − ��(�

��

����(����)

�� − � ���(�)��

��

=

= �����√� − �� � ���(����)

��

�� −�

����(�)�����√� − �� =

= � −�

����(�)�����√� − ��

� = −�

����(�)�����√� − ��; (�)

� = �������

�√� − ��

�� =⏞������

������(����)

��������������

��

= ������(����)

������

��

�(�) = ������(�����)

������

��

→ ��(�) = �������

����(� + �������)

��

= ���

� + �������

��

=

= ��������

����� + �������

��

= ��������

� + �����(� + ��)

��

=⏞

������

����������

= ���

� + ��√� + ����

=�

√� + ������� ���� + ����

=�

�√� + ��

Page 58: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

��(�) =�

�√� + ��→ �(�) =

��

��

√� + ��=⏞

����� � �

��

����� ��

��� �=

=�

�� ��� � �� =

���� �� + �� + ���

�(�) =�

�����√� + �� =

���� �

√� − �� = −

�����√� − ��; (�)

From (�), (�) it follows that:

�������

�√� − ��

�� = ����(√� − �) � ���(����)

��

��

Solution 3 by Rana Ranino-Setif-Algerie

� = �������

�√� − ��

�� = � ����

(� + ����)√� − ��

� ���

���

(� + ����)√� − ��

=⏞����� �

���

� + �������

��

=⏞������

���

� + ��(�� + �)

=�

���� + �

� =�

��

��

��� + �

=�

�������(�) =

�����√� + �� = −

�����√� − �� =

= −�

����� ⋅

����√� − ��

����

����� � ���(����)��

��

= −�

�����

Therefore,

�������

�√� − ��

�� = ����(√� − �) � ���(����)

��

��

Solution 4 by Samar Das-India

� = �������

�√� − ��

�� =⏞������

������(����)

��������������

��

= ������(����)��

����

��

�(�) = ������(�����)

������

��

Page 59: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

��

��= �

����

(� + �������)������

��

= ���

� + �������

��

=

= ��������

� + (� + ��)�����

��

=�

�� + ��

�������

��

√�� + ��

+ �����

��

=

=�

�� + �⋅

√�� + �

����� �����

√�� + �

��

=�

�√�� + �; (�)

→ � =�

��

��

√�� + �=

���� �� + ��� + �� + �; (�)

��� � = � → �(�) = ������(� ⋅ ����)

������

��

= � ���� (�):

�(�) =�

����(�) + � → � = �

�(�) = ������(� ⋅ ����)

������

��

=⏞(�) �

������ + √��; (�)

� = �����√� − �� � ���(����)

��

�� =⏞

�→��

��

�����√� − �� � ��� ���� ��

�− ��� (−��)

��

=

= �����√� − �� � ���(����)

��

��

�� = �����√� − �� � ���(���� ⋅ ����)

��

�� = �����√� − �� � ��� ������

��

��

�� =

= �����√� − ��(� ���(�����)

��

�� − � ���(�)

��

�� =⏞����

= �����√� − ��(�

��

����(����)

�� − � ���(�)��

��

=

= �����√� − �� � ���(����)

��

�� −�

����(�)�����√� − �� =

= � −�

����(�)�����√� − ��

Page 60: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

� = −�

����(�)�����√� − ��; (��)

From (�), (��) it follows that:

�������

�√� − ��

�� = ����(√� − �) � ���(����)

��

��

Solution 5 by Akerele Segun-Lagos-Nigeria

�����(�) = �(−�)������

�� + �

���

, |�| < 1

��, �������

�√� − ��

�� = � �(−�)������

(�� + �)��√� − ���

���

���

=

= �(−�)�

�� + �� ���(� − ��)�

��

���

�� =⏞���� �

��

(−�)�

�� + �� ���

��(� − �)�

����

���

=

=�

��

(−�)�

�� + �� �� +

�,�

��

���

=�

��

(−�)���(��)

(�� + �)�(� + �)�(� + �)

���

= −�

�����√� − �� =

= �����√� − �� �−�

����(�)�

������: � ���(����)��

��

= −�

�����

Therefore,

�������

�√� − ��

�� = ����(√� − �) � ���(����)

��

��

�� = � �����

�� + �∙

��

���

�������

�� + �+ ����

���

�� + ��

UP.362. Find:

� = ����→�

��

������(���)

���

Proposed by Florică Anastase-Romania

Page 61: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

Solution by proposer

∵ � ������

�� + ��

��

���

�������

�� + �+ ����

���

�� + �� = �� − �, ∀� ∈ ℕ

� ������

�� + ��

��

���

�������

�� + �+ ����

���

�� + �� =

= � ������

�� + ��

���

�������

�� + �+ ����

���

�� + �� + � ����

��

�� + ��

��

�����

�������

�� + �+ ����

���

�� + �� =

= �� + ��

Denote �� + � − � = �, how � ∈ {� + �, � + �, … , ��} ⇒ � ∈ {�, �, … , � − �, �}

�� = � ���(�� + � − �)�

�� + �

���

����(�� + � − ��)�

�� + �+ ����

(�� + � − ��)�

�� + �� =

= � �����

�� + �

���

�������

�� + �− ����

���

�� + �� ⇒

� = �� + �� = � � ������

�� + ����

���

�� + �

���

= � � ������

�� + �

���

=

= � � �� + ������

�� + ��

���

= �� + � � ������

�� + �

���

= �� − � ⇒ �� = �� − �

∵ ���[�� − (�� − �)]�

�� + �= −���

���

�� + �, ∀� = �, � … , �

∵ � ������

�� + �

���

= � − � ����(�� − �)�

�� + �+ ⋯ + ���

�� + �� = −

�;

Hence,

� = ����→�

��

������(���)

���

= ����→�

��

��������(���)

���

= ����→�

��

������

���

; (�)

������=

��

����=

��

�� + (� − �)���� ≤ �

� − ��

∙�

�����

�=

Page 62: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

= �� +�

� − �� �� +

� − ��

���

∙�

�����

�≤

��

�����

�⇒

� ≤ ��

������(���)

���

≤ � + �� ��

�����

���

���

+�

�≤ � + �� ∙

� − �

�����

�+

Therefore,

� = ����→�

��

������(���)

���

= �

UP.363 Let be (��)���; (��)��� ⊂ (�, ∞) such that:

����→�

(���� − ��) = � ∈ (�, ∞); �� = �� �����

���

��

Find:

� = ����→�

����� ⋅ ����

� + �−

�� ⋅ ��

��

Proposed by D.M. Bătineţu-Giurgiu, Daniel Sitaru-Romania

Solution 1 by proposer

By Cesaro-Stolz’s theorem:

����→�

��

�= ���

�→�

����

��= ���

�→�(����−��) = �

By Cauchy D’Alembert’s theorem:

����→�

��

�= ���

�→��

∏ ���������

��

= ����→�

∏ �����������

(� + �)���⋅

��

∏ ������

=

= ����→�

��

� + ��

⋅�����

� + �=

�⋅ ���

�→�

�����

�� + �=

�⋅ ���

�→�

�����

�� + �⋅

�� + �

� + �=

��

Hence,

����→�

�� ⋅ ��

��= ���

�→�

��

�⋅

��

�= � ⋅

��

�=

���

Denote �� =����⋅����⋅�

��⋅��⋅(���), hence ���

�→��� = �, ���

�→�

����

�����= �.

Page 63: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

����→�

��� = ���

�→��

� + ��

⋅∏ �����

������

∏ ���������

⋅�

�∏ �����������

���⋅ �

�����

���

=

=�

�⋅ ���

�→�

����

� + �⋅

� + �

����⋅ ��� +

���� − ��

���

���������

⋅�

��⋅(�������)

� =

=�

�⋅ �� ⋅

��⋅ ��⋅

�� = �

� = ����→�

����� ⋅ ����

� + �−

�� ⋅ ��

�� = ���

�→�

����

�⋅

�� − �

�����⋅ �� =

= ����→�

����

��⋅

�� − �

�����⋅ �����

� =���

�⋅ � ⋅ � =

���

Solution 2 by Amrit Awasthi-India

We are given that: ����→�

(���� − ��) = �. This can be rewritten as:

����→�

���� − ��

(� + �) − �= �

Using Cesaro-Stolz Theorem, we have:

����→�

(���� − ��) = ����→�

���� − ��

(� + �) − �= ���

�→�

��

�= � → �� = ��

Now substituting this value in the given value of �� we have

�� = ��(�� − �)�

���

��

= � ⋅ �� ⋅ � ⋅ � ⋅ … ⋅ (�� − �)���

Now, the product can be rewritten as

�� ⋅ � ⋅ � ⋅ … ⋅ (�� − �)� =(��)‼

�! ⋅ ��→ �� = � ⋅ �

(��)!

�! ⋅ ���

��

Now, we have:

� = ����→�

����� ⋅ ����

� + �−

�� ⋅ ��

�� = ���

�→�

���� ⋅ ����

� + � −�� ⋅ ��

�(� + �) − �

Therefore, again using Cesaro-Stolz theorem, we have:

Page 64: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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� = ����→�

�� ⋅ ��

��

Now, substituting previously obtained values, we have:

�����→�

�� ⋅ ��

��= ���

�→�

�� ⋅ �(��)!�! ⋅ ���

��

Therefore, using Stirling’s approximation as � approaches infinity, we have:

�(��)!��� ≈ �

��

��

⇔ (�!)�� ≈

Hence, we have:

� = ����→�

��

��⋅

���

��⋅

�=

���

Therefore,

� = ����→�

����� ⋅ ����

� + �−

�� ⋅ ��

�� =

���

UP.364 In ���� the following relationship holds:

���

����

� ���

��

���

� ⋅ … ⋅ ���

��

���

� ≥��

���√������

; � ∈ ℕ∗

Proposed by D.M. Bătineţu-Giurgiu, Daniel Sitaru-Romania

Solution 1 by proposer

��

��

���

= �����

��

���

≥����� (� + � + ⋯ + �)���

(� + � + �)�=

����

(��)�=

�√������

(��)�≥

����������

≥�√��

����

��√���� =

�√���������

��=

�√�����

��

� ���

�����

���

≥ ��√��

���

��

���

=�√��

(���)�(���)�⋯�(���)

�������⋯��=

�√�����

�(���)�

��(���)

=

Page 65: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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=�√��

��

��√���(���)

=��

���√������

.

Equality holds for an equilateral triangle.

Solution 2 by Adrian Popa-Romania

��

��

���

=����

��+

����

��+

����

��≥⏞

����� (� + � + �)���

(� + � + �)�=

����

(��)�≥⏞

����√�

≥����

�� ⋅��√�

� �

� =����

��√���� =

�� ⋅ �

��� ⋅ �� ⋅ ��

=�

��� ⋅ ��

���

����

� ���

��

���

� ⋅ … ⋅ ���

��

���

� ≥�

��� ⋅ ��

⋅�

��� ⋅ ��

⋅ … ⋅�

��� ⋅ ��

=

=��

�������⋯��

� ⋅ �������⋯��

=��

���(���)

� ⋅ ��(���)

≥��

���√������

Solution 3 by George Florin Șerban-Romania

��

����

≥√�

�; (��������`)

� ��

���

���

≥⏞������ �

���

����

≥�

��

√�

��

, … , � ��

���

���

≥⏞������ �

������

����

≥�

�����

√�

��

���

����

� ���

��

���

� ⋅ … ⋅ ���

��

���

� ≥

√�� ⋅ �

√�� �

⋅ … ⋅ �√�� �

� ⋅ �� ⋅ … ⋅ ����=

=√�

������⋯��

�������⋯�� ⋅ ��(���)

=�� ⋅ �

������

������ ⋅ �����

=

Page 66: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

=�� ⋅ �

������

�����

������=

��

������ ⋅ �����

=��

������ ⋅ �������

=��

���√������

UP.365 Let be �� = ∑ ����� ��

��������

��� ; � ≥ �. Find:

� = ����→�

��(����� − ���)

Proposed by D.M.Bătineţu-Giurgiu, Daniel Sitaru-Romania

Solution 1 by proposers

It’s easy to prove that:

����� ��

�� + � + �� = ����� �

�� − ����� �

� + �� , ∀� ∈ ℕ∗

It follows that:

�� = ������ − ����� ��

���� =

�− ����� �

����and ���

�→��� =

� = ����→�

��(����� − ���) = ����→�

�����(�������� − �) =

= ����→�

����� ���

(���)��(���)�� − �� = ����→�

����� ���

������� − �� =

= ����→�

����� ��

�������

��������

− �

����� ��

�� + �� + ��

� ⋅ ����� ��

�� + �� + �� =

= ����→�

��� ⋅�

������� − �

����� ��

�� + �� + ��

⋅����� �

��� + �� + �

��� + �� + �

⋅��

�� + �� + �=

= ��� ⋅ � ⋅ � ⋅ � = �

��

Solution 2 by Asmat Qatea-Afghanistan

�� = � ����� ��

�� + � + ��

���

= �(�����(� + �) − ����� �

���

) = �����(� + �) − ����� �

�� = �����(� + �) −�

Page 67: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

� = ����→�

�� �������(���)��� − ������(���)�

��� =

=�

√��� ⋅ ����→�

�� ������(���)����������(���)������(���) − �� =

= ����→�

������(���) ⋅�

�������

��������

− �

����� ��

�� + �� + ��

⋅����� �

��� + �� + �

���

=

= ��� ⋅ ���

�→�

−�� + �

(�� + �� + �)�

� + ��

�� + �� + ��

−�

��

=�

��

�⋅ ���

�→�

��� + ���

(�� + �� + �)� + �=

���

�⋅

�= �

��

Therefore,

� = ����→�

��(����� − ���) =�

√��� ⋅ ���� = √���

Solution 3 by Ravi Prakash-New Delhi-India

��� �� = ����� ��

�� + � + �� = ����� �

� + � − �

� + �(� + �)� = �����(� + �) − ����� �

⇒ � ��

���

= �����(� + �) −�

�= ��

��(����� − ���) =

= ������(���)��� ⋅

������(���) − ������(���)

�����(� + �) − �����(� + �)⋅

����� ��

� + (� + �)(� + �)�

�� + (� + �)(� + �)

⋅��

(� + �)(� + �) + �

� = ����→�

��(����� − ���) = ���

��� = �

��

Solution 4 by Kaushik Mahanta-Assam-India

�� = � ����� ��

�� + � + ��

���

= �(�����(� + �) − ����� �

���

) = �����(� + �) − ����� �

Page 68: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

�� = �����(� + �) −�

� = ����→�

�� �������(���)��� − ������(���)�

��� =

= ����→�

�������������(���) − ������(���)� =

= ����→�

������ ���

�� −

���

� + �+ ⋯ � − ��

�� −

���

� + �+ ⋯ �� =

= ����→�

������ �

���

� + �−

���

� + �� = ���

�→��

��

����� ⋅

�� �� +��� �� +

���

= ���

UP.366. If �� = � +�

√�+ ⋯ +

√�− �√�; � ≥ � find:

� = ����→�

(� + ����� − ���)�√�

Proposed by D.M.Bătineţu-Giurgiu, Daniel Sitaru-Romania

Solution 1 by proposers

It’s known that: ����→�

�� = � ∈ (−�, −�).

It follows that: ����→�

(����� − ���) = �� − �� = �

� = ����→�

((� + ����� − ���)�

��������

)�√�(���������) = ����

�→��√�(���������)

; (�)

����→�

�√�(����� − ���) = ����→�

���(�������� − �)�√� =

= �� ⋅ ����→�

�������� − �

���� − ��⋅ �√�(���� − ��) =

= �� ⋅ � ⋅ ����→�

�√� ��

√� + �− �√� + � + �√�� =

= �� ⋅ ����→�

�√� ��

√� + �− ��√� + � − √��� = �� ���

�→��√� �

√� + �−

√� + � + √�� =

= �� ����→�

�√� ⋅√� + � + √� − �√� + �

√� + ��√� + � + √��=

Page 69: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

= −�� ����→�

√� + ��√� + � − √������→�

√��√� + � − √�� =

= −�

��� ���

�→�

√�

√� + � + √�= −

���; (�)

From (1),(2) it follows that:

� = ����→�

(� + ����� − ���)�√� = ����

Solution 2 by Marian Ursărescu-Romania

� = ����→�

((� + ����� − ���)�

��������

)�√�(���������) = ����

�→��√�(���������)

; (�)

����→�

�√�(����� − ���) = ����→�

���(�������� − �)�√� =

= �� ⋅ ����→�

�������� − �

���� − ��⋅ �√�(���� − ��); (�)

����→�

��� = ��, ����� � − �� ����������� ��������; (�)

����→�

�������� − �

���� − ��= ��� � = �; (�)

����→�

�√�(���� − ��) = ����→�

�√� ��

√� + �− �√� + � + �√�� =

= ����→�

�√� ��

√� + �− ��√� + � − √��� = ���

�→��√� �

√� + �−

√� + � + √�� =

= ����→�

�√� ⋅√� + � + √� − �√� + �

√� + ��√� + � + √��=

= −����→�

√� + ��√� + � − √������→�

√��√� + � − √�� =

= −�

�����→�

√�

√� + � + √�= −

�; (�)

From (�), (�), (�), (�), (�) it follows that: � = ����→�(� + ����� − ���)�√� =�

�����

Solution 3 by Felix Marin-Romania

�� = � +�

√�+ ⋯ +

√�− �√�.

Page 70: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

Indeed, �� is related to the zeta function. Namely,

�� = � ��

�� +

��

{�}

���

���

����������

��

√�

���� ���� ����→�

�� = � ��

��

Moreover,

� + ����� − ��� = � + ���(�������� − �) ~⏞�� �→�

� + ���

��

�(���� − ��) =

= � + ���

��

��

√� + �− �√� + � + �√��

~⏞�� �→�

� + ���

��

��

���

�� −�

��� − ��

�� �� +

��−

���� + ��

��� = � −

���

��

�−

�√�

Therefore,

� = ����→�

(� + ����� − ���)�√� = ����→�

⎩⎪⎨

⎪⎧

� −

⎣⎢⎢⎢⎡

���

�� �����

⎦⎥⎥⎥⎤

⋅�

�√�

⎭⎪⎬

⎪⎫

�√�

=

= ���

⎜⎛

�� �����

⎟⎞

≈ �. ����

UP.367. Find: � = ����→�

��(��)‼�

(���)‼

; � ∈ ℕ

�‼ = �, (��)‼ = � ⋅ � ⋅ … ⋅ (��), � ∈ ℕ∗

Proposed by D.M. Bătineţu-Giurgiu and Daniel Sitaru-Romania

Solution by proposers

Denote �� = ��(��)‼�

(���)‼

; � ≥ �; �� =(��)‼

��

For � = �; �� = �(�‼)�

�‼

�= �; � ≥ �; ���

�→��� = �

Page 71: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

If � = �; �� = ��(��)‼�

(���)‼

= �; � ≥ �; ����→�

�� = �

For � ≥ �: �� = ��(��)‼�

(���)‼

= ��(��)‼

��

⋅��

�(���)‼� = � �

(��)‼

��

��

⋅ ����

(���)‼

����→�

�� = ����→�

��(�� + �)‼

(� + �)���⋅

��

(��)‼�

⋅(� + �)�(���)

(��� + ��)‼⋅

(���)‼

���� =

= ����→�

���� + �

� + �⋅

���

⋅ ��� ⋅

(� + �)�

(��� + �)(��� + �) ⋅ … ⋅ (��� + ��)� =

=��

��⋅ �� ⋅ ���

�→�

(� + �)�

��(�� + �)(�� + �) ⋅ … ⋅ (�� + �)=

= ����→�

�� + �

�� + �⋅

� + �

�� + �⋅ … ⋅

� + �

�� + �� =

��

UP.368. Let be

�� = ���

�� + � +��!

+��

�!+ ⋯ +

��

�!

�����

; � ∈ ℕ∗

Prove that: �� < �!

Proposed by D.M. Bătineţu-Giurgiu,Daniel Sitaru-Romania

Solution 1 by proposers

Let be ��(�) = � +�

�!+

��

�!+ ⋯ +

��

�!

��� (�) = � +

�!+

��

�!+ ⋯ +

����

(� − �)!= ����(�)

��(�) − ��� (�) =

��

�!⇒ �� = �! ���(�) − ��

� (�)�

�� = �! ���(�) − ����(�)�

��(�) = ��! ���(�) − ��

� (�)�

�� + ��(�)��

���

= �! ��! ��� + ��(�) − �� − ��

� (�)�

�� + ��(�)��

���

=

�! � �����

− �! ��� + ��

� (�)

�� + ��(�)��

���

= �! − �! ������ + ��(�)��

���=

Page 72: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

= �! �� − ������� + ��(� + �)

�� + ��(�)� < �!

Because ���� > ��; ��(� + �) > ��(�), �� −increasing function, hence

�������(���)

�����(�)> �and ���

�������(���)

�����(�)> �.

Solution 2 by Ravi Prakash-New Delhi-India

For � ≤ � ≤ � + � we have:

�� + � +�

�!+

��

�!+ ⋯ +

��

�!> �� →

�(�) =��

�� + � +��! +

��

�! + ⋯ +��

�!

<��

��

→ � �(�)�����

≤ � ��������

�� < � ������

��

Therefore,

� �(�)���

�� < �(� + �) = �!

Solution 3 by Jaihon Obaidullah-Afghanistan

For � ≤ � ≤ � + � we have:

�� + � +�

�!+

��

�!+ ⋯ +

��

�!> �� →

Also this integral smaller then

� ��������

= � �(���)������

�� = �(� + �) = �!

Solution 4 by Remus Florin Stanca-Romania

���

�� + � +��! +

��

�! + ⋯ +��

�!

�����

< �! ⇔

��

�!

�� + � +��! +

��

�! + ⋯ +��

�!

�����

< �

Page 73: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

Let ��(�) = � +�

�!+

��

�!+ ⋯ +

��

�! then,

���(�) − ����(�)

�� + ��(�)��

���

< � ⇔ ������(�) − �������(�)

� + �����(�)��

���

< �

��� (�) = � +

�!+ ⋯ +

����

(� − �)!= ����(�)

⇔ ������(�) − �����

� (�)

� + �����(�)��

���

< � ⇔ − �−�����(�) + �����

� (�)

� + �����(�)��

���

< �

− �(���)���(�) + �����

� (�)

� + �����(�)��

���

< � ⇔ − �(�����(�) + �)�

�����(�) + ���

���

< �

⇔ − ���(�����(�) + �)|���� < �

⇔ ��� ������(�) + �

�������(� + �) + �� < � ⇔

�����(�) + �

�������(� + �) + �< �

⇔ � − � + ������(� + �) − ��(�)� > �; (�)

��� (�) = ����(�), ∀� ≥ � → ��(�) −increasing for � ∈ ℝ� → ��(� + �) > ��(�)

→ ������(� + �) − ��(�)� ≥ � and � − � > � then

� − � + ������(� + �) − ��(�)� > � and from (�) it follows that

���

�� + � +��! +

��

�! + ⋯ +��

�!

�����

< �!

UP.369. Let �: ℝ → ℝ be a continuous function; �, � > �; � < �; � + � = �;

�(� − �) + �(�) = �; ∀� ∈ ℝ; � > �. Find:

� = � (�� − �� + ��)�(�)���

Proposed by D.M. Bătineţu-Giurgiu and Daniel Sitaru-Romania

Solution 1 by proposers

� = �(�) = � − �; ��(�) = −�; �(�) = �; �(�) = �

� = � (�� − �� + ��)�(�)���

= � ((� − �)� − �(� − �) + ��)�(� − �)���

=

Page 74: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

= � (�� − �� + ��)�� − �(�)����

= � � (�� − �� + ��)���

− � (�� − �� + ��)�(�)���

� =�

�� (�� − �� + ��)��

=�

��

�� − ��

�−

�� − ��

�+ ��(� − �)� =

=�(� − �)

��(��� − ��� + ���) =

�(� − �)(��� − ���)

��

Solution 2 by Ravi Prakash-New Delhi-India

�����: � �(�)���

= � �(� + � − �)���

, �� ���:

� = � [(� − �)� − �(� − �) + ��]�(� − �)���

= � (�� − �� + ��)�� − �(�)����

⇒ �� = � � (�� − �� + ��)���

= � ���(� − �) −�

�(�� − ��) +

�(�� − ��)� =

= �(� − �) ��� −�

�⋅ � +

�(�� + �� + ��)� =

�[��� − �(�� + �� + ��)](� − �) =

=�

�[��� − �(� + �)� + ���](� − �) =

�(�� + ���)(� − �)

Solution 3 by Kamel Gandouli Rezgui-Tunisia

� (�� − �� + ��)�(�)���

=⏞�����

� ((� − �)� − �(� − �) + ��)�(� − �)���

=

� (−�� + �� + ��)�(� − �)���

= � (−�� + �� + ��)�� − �(�)����

⇒ � � (�� − �� + ��)�(�)���

= � � (−�� + �� + ��)���

=

= � �−���

�+

��

�+ ����

= −����

�+

���

�+ ���� +

����

�+

���

�+ ���� =

= −����

�+

���

�+ ��� +

����

�+

���

Solution 4 by Kaushik Mahanta-Assam-India

� = � (�� − (� + �)� + (� + �)�)�(�)���

=

Page 75: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

= � �(� + �) − ���

− (� + �) �(� + � − �) + (� + �)�� �(� + � − �)���

=

= � ((� + �)� + �� − �(� + �)� − (� + �)� + (� + �)� + (� + �)�)�� − �(�)����

=

= � (�� − (� + �)� + (� + �)�)� ���

− �

⇒ �� = � (�� − (� + �)� + (� + �)�)��

�� = ���

�− (� + �)

��

�+ (� + �)����

⋅ � =

� =�

�((�� − ��)

�− (� + �) ⋅

�� − ��

�+ (� + �)�(� − �) =

�[(� + �)� + ���](� − �)

Solution 5 by Angel Plaza-Spain

By doing � = � − �, since �� − �� + �� = (� − �)� − �(� − �) + �� = �� − �� + ��and

�(�) = �(� − �) = � − �(�), the integral becomes

� = � (�� − �� + ��)�� − �(�)��

�� = −� + � � (�� − �� + ��)�

��

→ � =�

�� (�� − �� + ��)

�� =�(� − �)

��(��� + ��� + ���)

UP. 370.If �, � > �, then

���

(� + �)(������� + �������)

��

<�

��(� + �)��

�+ ������ �

���

Proposed by Florică Anastase-Romania

Solution:

Theorem (Bonnet-Weierstrass):

If �: [�, �] → � decreasing function of ��class and �: [�, �] → � continuous function, then

∃� ∈ [�, �] such that:

� �(�)�(�)�� = �(�) � �(�)�� + �(�) � �(�)��

Page 76: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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Demonstration:

Let �: [�, �] → �, �(�) = �(�) − �(�)���������� ��� �(�) ≥ �, ∀� ∈ [�, �].

From second M.V.T. ∃� ∈ [�, �] such that:

� �(�)�(�)�� = �(�) � �(�)��

� �(�)��(�) − �(�)��� = (�(�) − �(�)) � �(�)��

� �(�)�(�)��

=

= �(�) � �(�)�� + (�(�) − �(�)) � �(�)�� =

�(�) � �(�)�� + �(�) � �(�)��

q.e.d.

Let �, �: ��,�

�� → �, �(�) =

���������������, �(�) =

���, ��(�) = −

(���)� < � then f is

decreasing

�(�) = ���

������� + �������=

= ��

�� + �������∙

��

�����=

���

��

���

��

+ ��

=�

������� �

�����

�� + �

Then ∃� ∈ ��,�

�� for which:

���

(� + �)(������� + �������)

��

= �(�)��(�) − �(�)� + � ��

�� ��(�) − �(�)� =

=�

������� �

�����

�� +

��� + �

∙�

��������

�− ����� �

������� =

=�

��(� + �)������� �

������ + ������

��

Page 77: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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∵ ������ < �, ∀� > � → ����� ��

������ <

���� � <

����

�=

Therefore,

���

(� + �)(������� + �������)

��

=�

��(� + �)������� �

������ + ������ �

��� <

<�

��(� + �)��

�+ ������ �

���

UP.371. Let be (��)���; �� = ∏ �(�� − �)‼��

����� . Find:

� = ����→�

�(� + �)�

��������

−��

����

Proposed by D.M. Bătineţu-Giurgiu, Daniel Sitaru-Romania Solution by proposers

����→�

����

= ����→�

���

��

= ����→�

(� + �)���

����⋅

��

��= ���

�→��

� + �

��

⋅� + �

�(� + �)‼���

=

= �����→�

�(�� − �)‼�

= �����→�

���

(�� − �)‼

= �����→�

(� + �)���

(�� + �)‼⋅

(�� − �)‼

��=

= ����→�

�� + �

��

⋅� + �

�� + �=

��

�.

If �� = ����

��

⋅���

�������� ; ∀� ≥ � then:

����→�

�� = � and ����→�

����

�����= �

����→�

��� = ���

�→��

� + �

��

��

⋅��

����⋅ �����

��� = �� ����→�

�(�� + �)‼���

⋅�����

���

� + �= �� ⋅

�⋅

��

= �

Hence,

� = ����→�

�(� + �)�

��������

−��

����

� = ����→�

��

����

(�� − �) = ����→�

��

����

⋅�� − �

�����⋅ �� =

Page 78: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

= ����→�

����

⋅�� − �

�����⋅ ��

� =��

�⋅ � ⋅ ���� =

��

UP.372. (�)��� −be a positive sequence of real numbers such that �� > �

and

���� =��

��� − ���(� + ��)

Find:

� = ����→�

(��)‼

(�� − �)‼⋅

���

√�� + �

Proposed by Florică Anastase-Romania

Solution 1 by proposer

Let be the function: �: (�, ∞) → ℝ, �(�) =��

������(���)

�� > �. Suppose �� > � and how �� > ���(� + ��) ⇒ ��� > ���(� + ��) ⇒

���

��� − ���(� + ��)> � ⇒ ���� > �

�� > ���(� + ��) ⇒ �� − ���(� + ��) > � ⇒ ��� − ���(� + ��) > �� ⇒����

��< �

⇒ (��)��� ↘. So, (��)��� −is convergent sequence.

����→�

���= ���

�→��

����−

��� = ���

�→�

�� − �(��)

���(��)= ���

�→�

� −��

�� − ���(� + �)

��

�� − ���(� + �)

=

= ����→�

�� − ����(� + �)

��= ���

�→�

�(� + �)�

�=

Therefore, ����→�

��� = �; (�)

��(�) = ���

√� − ����

; � ∈ ℕ ⇒

��(�) = −���� ⋅ �� − ����

+ (� − �) � ������ − �����

=

Page 79: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

= (� − �)���� − (� − �)�� − ������ − �� ⇒

�� =� − �

�⋅ ���� = ⋯ =

(� − �)!

�!��

��� =(�� − �)‼

(��)‼⋅

�; ����� =

(��)‼

(�� + �)‼

How ����

����→ �, then

�����

���→ � and ���

�→��

(��)‼

(����)‼�

⋅�

����=

�; (�)

From (1),(2) it follows that:

� = ����→�

(��)‼

(�� − �)‼⋅

���

√�� + �= ��

�= √��

Solution 2 by Adrian Popa-Romania

(��)‼

(�� − �)‼=

� ⋅ � ⋅ � ⋅ … ⋅ (��)

� ⋅ � ⋅ � ⋅ … ⋅ (�� − �)=

�� ⋅ �!

(��)!�� ⋅ �!

=�� ⋅ (�!)�

(��)!=

��

(��)!(�!)�

=��

����

����→�

����

�� √�� = �; (������)

����

��≅

√��→

��

����

�≅ √��

����→�

(��)‼

(�� − �)‼⋅

√�� + �= ���

�→�

√��

√�� + �= ���

�→��

��

�� + �=

���� =��

��� − ���(� + ��) ��� ���(� + ��) ≤ �� →

���� <��

��� − ��=

���

��= �� → (��)��� ↘

Let be � = ����→�

�� → � =��

������(���)→ �� = � ���(� + �) → �(� − ���(� + �)) = � → � = �

So, ����→�

�� = �.

Now, we have:

���� =��

��� − ���(� + ��)=

���

��� − �� +��

� −��

� +��

� − ⋯=

Page 80: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

=�

���

−�

��+

�� −

��

� +��

� − ⋯����������

=�

���

+��

=���

� + ��

�� =���

� + ��, �� =

���

� + ��=

���

�� + ����

� + ��+ �

=���

� + ��⋅

� + ��

� + ���→

�� =���

� + ���

�� =���

� + ��=

���

� + ���

���

� + ���+ �

=���

� + ���⋅

� + ���

� + ���=

���

� + ���

Applying principle of mathematical induction, we get:

�� =���

� + (� − �)��→ ���

�→�

����

��� − �� + �= �

Therefore,

� = ����→�

(��)‼

(�� − �)‼⋅

���

√�� + �= ��

�= √��

UP.373 � ∈ ℕ, � > � and �� > �, ���� =��

����; � ∈ ℕ∗

Find:

� = ����→�

��

���(��� �)⋅

���� − �

��

Proposed by Florică Anastase-Romania

Solution 1 by Ruxandra Daniela Tonilă-Romania

���� =��

�� − �⇔

����

��=

��

�� − �> � ⇒ ���� > ��, �� (��)��� ↗

���� =��

�� − �|(−��) ⇒ ���� − �� =

���

�� − �

Page 81: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

⎩⎪⎪⎨

⎪⎪⎧ �� − ���� =

�����

���� − �

���� − ���� =�����

���� − �⋮

�� − �� =���

�� − �

Adding these relations, we get:

�� = �� +���

�� − �+

���

�� − �+ ⋯ +

�����

���� − �=

= �� + � ���

�� − �+

��

�� − �+ ⋯ +

����

���� − �� > � ⋅ �

Therefore,

����→�

�� = ∞

����→�

���� =⏞

���

����→�

����

��= ���

�→�

� +�� +

�� + ⋯ +

�� + �

� +�� +

�� + ⋯ +

��

= �

� = ����→�

��

���(����)⋅

���� − �

��= ���

�→�

��⋅

���� ����

− �

��� ����

⋅ ��� ���� ⋅

���(��� �)=

= ����→�

��⋅ ���

�→�

���(��� �)⋅ ��� ���

� ; (�)

����→�

��=⏞

���

����→�

� + � − �

���� − ��= ���

�→�

���

�� − �− ��

= ����→�

�� − �

���=

�; (�)

����→�

���(��� �)⋅ ��� ���

� = ����→�

��� ��

���(��� �)= ���

�→�

��� �� − ���(��� �) + ���(��� �)

���(��� �)=

= ����→�

�� +��� �

��

��� ��

���(��� �)� = ���

�→��� +

��� ��� − ��� �

��� �+ ��

���(��� �)� = �; (�)

∵ ����→�

(�� − ��� �) = � ��� ����→�

��� ��� − ��� �

��� �+ �� = �

From (�), (�), (�) it follows that:

Page 82: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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�� NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS

� = ����→�

��

���(����)⋅

���� − �

��=

Solution 2 by proposer

Let �� =�

��; �� <

�⇒ �� ∈ ��,

�� , ���� = ��(� − ���)

Suppose �� ∈ ��,�

�� ⇒ � < ��� < � ⇒ � < � − ��� < � ⇒ � < ��(� − ���) < �� ⇒

���� ∈ ��,�

�� and (��)��� ↘. So, (��)��� −convergent sequence, then ∃� ∈ ℝ such that

����→�

�� = � ⇒ � = �(� − ��); �� <�

�� ⇒ � = �

����→�

���= ���

�→�

���

�= ���

�→�

�� − ����

�� ⋅ ����= ���

�→�

�� − ��(� − ���)

���(� − ���)

=

= ����→�

����

���(� − ���)

= ����→�

� − ���= �

Thus, ����→�

��=

�� = ���� ; � < �� < √�

�→ � ⇒ �� − � → �

� = ����→�

��

���(��� �)⋅

���� − �

��= ���

�→�

��⋅

�� ���� − ��

���(��� �)=

=�

�⋅ ���

�→�

�� − �

�� ��� �� +

�� +

�� + ⋯ +

���

⋅��� �� +

�� +

�� + ⋯ +

���

���(��� �)=

=�

�⋅ ���

�→�

�� − �

�����⋅

⎜⎜⎜⎛��� �

� +�� +

�� + ⋯ +

��

��� ��

���(��� �)+ �

⎟⎟⎟⎞

=

=�

�⋅ ���

�→�

���[� + (�� − �)]�

����

⎜⎜⎜⎛��� �

� +�� +

�� + ⋯ +

��

�����

���(��� �)+ �

⎟⎟⎟⎞

=�

Page 83: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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UP.374 Calculate the integral:

�� �� �

�� + �√� + �

��

Proposed by Vasile Mircea Popa – Romania

Solution 1 by proposer

Let us denote: � = ∫� �� �

����√���

���

In this integral we make the variable change: � = ��

We obtain:

� =�

��

��� �� �

�� + � + �

��

We have, successively:

� =�

��

(� − �)��� �� �

� − ��

��, � =�

���

��� �� �

� − ��

�� − ��

�� �� �

� − ��

���

� =�

��� � ����

��

���

�� � �� − � � ������

���

�� �

���

� =�

�� �� ����

��

�� � �� − � ������

�� � ���

���

We will use the following relationship:

∫ �� �� � ���

�= −

(���)�, where � ∈ ℝ, � ≥ �

We obtain:

� =�

�� �

��� +���

� −�

��� +���

��

���

Or:

Page 84: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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� =�

�� �

��

�� +���

� −

��

�� +���

� �

���

We now use the following relationship:

��(�) = ��

(� + �)�

���

where ��(�) is the trigamma function.

We have obtained the value of the integral required in the problem statement:

� =�

����� �

�� − �� �

���

Solution 2 by Kaushik Mahanta-Assam-India

� = �� �� �

�� + �√� + �

�� =⏞√���

��������

�� + �� + �⋅ ����

= ��� ��� ��

(��)� + �� + �⋅ ����

=

= ���� ��� � (� − ��)

(� − ��)(� + �� + ��)⋅ ����

= �� ���(� − ��) ��� �

� − ��

�� =⏞����

= � ��

�� �� − �

��� ��� �

��

� − �⋅

��

����

=�

���

���� �� − �

��� ��� � ��

� − �

=

=�

����

���� ��� �

� − ���

− ���

�� ��� �

� − ���

� =

=�

����

���

�� ��� �

� − ���

− ��

��

�� ��� �

� − ���

� =

=�

����(�) �

�� − �(�) �

���

Solution 3 by Serlea Kabay-Liberia

� = �� �� �

�� + �√� + �

�� =⏞����

� ��� ��� �

�� + �� + ���

= � �(�� − �)�� ��� �

�� − ���

Page 85: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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��� �(�) = ���

�� − ���

���� ���� �(�) = � ���(�)

��|��� −

��(�)

��|����

�(�) = − � � ��������

���

= − ��

� + �� + �

���

⇒��(�)

��= �

(� + �� + �)�

���

���, � = �( ��

(�� + �)�

�����

���

− ��

(�� + �)�

���

� = � ��

���(�) �

�� −

���(�) �

��� =

����(�) �

�� − �(�) �

���

Therefore,

� = �� �� �

�� + �√� + �

�� =�

����(�) �

�� − �(�) �

���

Solution 4 by Ajentunmobi Abdulquyyom-Nigeria

� = �� �� �

�� + �√� + �

�� =⏞√���

��� ��� ��

�� + �� + �⋅ ����

= � ��� ��� �

(��)� + �� + �

�� =

= � ���(� − ��) ��� �

� − ����

= � ��� ��� �

� − ����

�������������

− � ��� ��� �

� − ����

�������������

� = � � ����� ��� � ��

���

=⏞���

− � ��������

�� + ���

= − �� � ⋅�����

(�� + �)�

���

=

= −�

���

�� +���

���

= −�

���(�) �

��

� = � ��� ��� �

� − ����

= � � ����� ��� � ��

���

=⏞���

− � ��������

�� + ���

= − �� � ⋅�����

(�� + �)�

���

= −�

���

�� +���

���

= −�

���(�) �

��

Thus,

Page 86: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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� = � − � =�

����(�) �

�� − �(�) �

���

Solution 5 by Mohammad Rostami-Afghanistan

� = �� �� �

�� + �√� + �

�� =⏞√���

��� ��� ��

�� + �� + �⋅ ����

= � ��� ��� �

�� + �� + �

�� =⏞����

= � �� ��� √�

�� + � + ��

��

�√��� ��

=�

��

��� ��� �

�� + � + ���

=�

��

���(� − �) ��� �

� − ����

=

=�

��� �

�� �(��)�

��|���

���

���� − � ��� �(��)�

��|�������

���

� =

=�

���

��|��� � �

��

��������

���

��

��|��� � �

��

��������

���

� =

=�

��� �

� + �� + ���

���

��

���

− � ��

� + �� + ���

���

��

���

� =

=�

��� �−

(� + �� + ��)��

���

���

− � �−�

(� + �� + ��)��

���

���

� =

=�

��−

��

�� +���

���

+�

��

�� +���

���

� =

=�

����

�� +���

���

− ��

�� +���

���

� =�

����(�) �

�� − �(�) �

���

UP.375 In any convex polygon ���� … ��, � ≥ � with the area � and the

sides lengths ������ = ��, � = �, �, ���� = �� the following inequality

holds:

���� − ������� + ������

���

≥ �� ⋅ ����

Proposed by D.M. Bătinețu-Giurgiu – Romania

Page 87: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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Solution 1 by proposer

We have:

��� − ��� + ���

≥ �� + ��, ∀�, � ∈ ℝ�∗ = (�, ∞) (1)

Indeed, let be � ∈ ℝ�∗ , such that � = ��� then the relationship (1) becomes:

� ���� − ���� ⋅ � + ���

≥ ���� + �� ⇔ ���(�� − � + �)� ≥ (�� + �)�� ⇔

⇔ �(�� − � + �)� ≥ �� + � ⇔ ��� + ��� + � − ��� + ��� − �� ≥ �� + � ⇔

⇔ �� − ��� + ��� − �� + � ≥ � ⇔ (� − �)� ≥ � which is obvious with equality ⇔

⇔ � = � ⇔ � = �.

Hence, according with (1) we have:

���� − ������� + ������

���

≥�

�⋅ ����

� + ����� �

���

=

E. Just-N. Schaunmberger

= � ���

���

≥ � ⋅ � ⋅ ����

Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco

��� � �� ��� ������������� �� ��� ������� ���� … ��.

�� ���� ∶ �� − ������� + ���� ≥⏞�����

�� −�� + ����

�+ ���� =

�� + ����

�,

∀� = �, �,

→ ���� − ������� + ������

���

≥ � ��� + ����

��

��

���

≥⏞��� �

���

�� + ����

���

=�

��� ��

���

→ ���� − ������� + ������

���

≥���

�.

�� ���� ����, ��� ��� ������ �������� ���� � �����, �� ≥ ��. ����

���������,

���� − ������� + ������

���

≥ �� ⋅ ����

�.

Page 88: ROMANIAN MATHEMATICAL MAGAZINE SOLUTIONS · 2021. 1. 12. · Solution 2 by Mohamed Amine Ben Ajiba-Tanger-Morocco. . NUMBER 25-RMM SUMMER EDITION 2022-SOLUTIONS. SP.375 If ,, ∈

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It’s nice to be important but more important it’s to be nice.

At this paper works a TEAM.

This is RMM TEAM.

To be continued!

Daniel Sitaru