RMM INEQUALITIES MARATHON 1 – 100 · Shivam Sharma - New Delhi – India . Solutions by Daniel...

www.ssmrmh.ro

RMM

INEQUALITIES

MARATHON

1 – 100

www.ssmrmh.ro

Proposed by Daniel Sitaru – Romania

D.M.Batinetu-Giurgiu-Romania

Neculai Stanciu-Romania

Rustem Zeynalov – Baku – Azerbaidian

Nguyen Ngoc Tu – HaGiang – Vietnam

Sladjan Stankovic – Skopje

Nguyen Viet Hung – Hanoi – Vietnam

Ibrahim Abdulazeez – Zaria – Nigeria

Ilkin Guliyev-Azerbaidian

Mihály Bencze-Romania

Adil Abdullayev – Baku – Azerbaidian

Hoang Le Nhat Tung – Hanoi – Vietnam

Madan Beniwal – Varanasi – India

Shivam Sharma - New Delhi – India

www.ssmrmh.ro

Solutions by Daniel Sitaru – Romania

Mihalcea Andrei Ștefan – Romania, Soumitra Mandal-Chandar Nagore-India

Uche Eliezer Okeke-Anambra-Nigerie, Soumava Chakraborty-Kolkata-India

Anas Adlany - El Zemamra-Morocco, Antonis Anastasiadis – Katerini – Greece

Kevin Soto Palacios – Huarmey – Peru, Șerban George Florin – Romania

Fotini Kaldi – Greece, Ravi Prakash - New Delhi – India

Myagmarsuren Yadamsuren – Darkhan – Mongolia,

Redwane El Mellas – Casablanca – Morocco, Richdad Phuc – Hanoi – Vietnam

Nirapada Pal – Jhargram – India, Chris Kyriazis – Greece

Rustem Zeynalov – Baku – Azerbaidian, Leonard Giugiuc – Romania

Dat Vo-Quynh Luu – Vietnam, Khanh Hung Vu – Ho Chi Minh – Vietnam

Seyran Ibrahimov-Maasilli-Azerbaidian, Nguyen Thanh Nho-Tra Vinh-Vietnam

Nguyen Ngoc Tu-Ha Giang – Vietnam, Abdallah El Farissi – Bechar – Algerie

SK Rejuan-West Bengal – India, Geanina Tudose – Romania

Tuk Zaya-Ulaanbaatar-Mongolia, Abdul Aziz-Semarang-Indonesia

Abdelhak Maoukouf-Casablanca-Morocco, Abdallah Almalih-Damascus-Syria

Pham Quoc Sang – Ho Chi Minh – Vietnam, Le Khanh Sy - Long An – Vietnam

Marian Cucoanes – Romania, Ngo Minh Ngoc Bao – Vietnam

Marian Dincă – Romania, Sanong Hauyrerai-Nakonpathom – Thailand,

Sladjan Stankovic – Skopje,Rovshan Pirguliyev – Sumgait – Azerbaidian,

Rozeta Atanasova – Skopje,Boris Colakovic-Belgrade – Serbia

www.ssmrmh.ro

Bedri Hajrizi – Mitrovica – Kosovo,Nikola Djurici – Serbia

Togrul Ehmedov – Baku – Azerbaidian, Ngoc Minh Ngoc Bao - Gia Lang –

Vietnam, Le Minh Cuong - Ho Chi Minh – Vietnam

Hoang Le Nhat Tung – Hanoi – Vietnam, Khalef Ruhemi – Iordania

Mohammed Hijazi – Iordania, Imad Zak – Saida – Lebanon

Subhajit Chattopadhyay - Visva Bharati – India

www.ssmrmh.ro

1. From the book: “Math Phenomenon”

If < ≤ ≤ then:

( + )( + )( + ) ≤ ∏ ≤ ( + )( + )( + )


Solution by Mihalcea Andrei Ștefan – Romania

+ ≤ ⇔ ≤ ⇔ ≤ true

≤ + ⇔ ≤ ⇔ ≤ true

Multiplying the analogs ⇒ q.e.d.


If ≥ ≥ > 0, + + = 10 then:

+ + ≤+ +

+≤ + +


Solution 1 by Soumitra Mandal-Chandar Nagore-India ,

Solution 2 by Uche Eliezer Okeke-Anambra-Nigerie

Solution 1 by Soumitra Mandal-Chandar Nagore-India

+ ++

≥ ( + )

⎣⎢⎢⎢⎢⎡ ∵ + + ≥ ( + ) ,

+ + ≥ ( + )

+ + ≥ ( + ) ⎦⎥⎥⎥⎥⎤

= ( + + ). We need to prove,

( + + ) ≥ + + ( + + ) ⇔ ≥ , which is true. So,

www.ssmrmh.ro

∑ ≥ + + We need to prove,

+ + ≥+ +

+⇔ + + ≥

+ ++

⇔ + −( + + )

++ + −

( + + )+

+

+ + −( + + )

+≥

⇔ ( ) + ( ) + ( ) ≥ , which is true ∵ ≥ ≥

∴ + + ≥+ +

+≥ + +

Solution 2 by Uche Eliezer Okeke-Anambra-Nigerie

Left 1) + + ≤ ∑

⇒ + + ≤( + ) −

+

⇒ + + + + ≤ ( + )

⇒ + + + + ≤ + + +

To show ⇒ ∑ ≤ + - Goal. Now,

∑ ≤ ∑ ( ) = ∑( + ) = + + ≤ + (True)

Right 2) ∑ ≤ + +

⇒( + )( + ) ≤

++ + +

www.ssmrmh.ro

+ + + ≤+

+ + +

To show + ≤ ∑ - Goal. Now,

∑ = + + (*)

Using: + ≤ ; + ≤ ; + ≤

(*) ≥ + + = + ⇒ + ≥ ⋅ ∑ (True)

3. From the book “Math Phenomenon”

If , , , ∈ ( ,∞) then:

+ + ≤ ( + )( + )


Solution by Soumava Chakraborty-Kolkata-India

To prove

⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ≤ + + +

⇔ + + + ≤ + + +

⇔ + ≤ + (1)

Let = , = , = , = → , , , > 0 ∵ , , , > 0

∴ + − −

= + − −

= − − − = ( ) − ( ) ( − )

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

= −( − ) ( ) ≤ (∵ > 0, as , , , > 0)

www.ssmrmh.ro

equality at = ⇒ = ⇒ =

∴ of (1) – RHS of (1) ≤

⇒ (1) is true. Hence proved. (equality at = )

4. If , > 0, + + = 0 then:

+ + ≥ ( + ) √ +


Solution 1 by Anas Adlany-El Zemamra-Morocco, Solution 2 by Antonis

Anastasiadis-Katerini-Greece

Solution 1 by Anas Adlany-El Zemamra-Morocco

First, note that: ( + + ) = + + + ( + )( + )( + )( + + + + + )

Then when + + = , we will have the following

+ + = ( + + + + + );

Now, let’s go back to the main problem and check out what we are really

dealing with the problem asks us to show that:

+ + ≥ ( + ) √ + whenever , > 0

We have + + ≥ ( + ) √ +

⇔ ( + + + + + ) ≥ ( + ) √ +

⇔ ( + + )( + + + + + ) ≥ ( + ) √ +

and the last step can be explained as follows:

+ + = ⇒ ( + ) + =

⇒ + + + ( + ) = ⇒ + + = ( + = − )

Now, since , > 0; by the AM-GM inequality we get:

+ + ≥ √ + − √ + (1)

www.ssmrmh.ro

Also, ∑( + ) − − = + + + +

= ( + ) + + ( + )( + = − )

= (( + ) − ) − = − − = − = ( − ) ≥

Which prove that ∑( + ) ≥ + (2)

Finaly, from results (1) & (2) the proof is completed.

Solution 2 by Antonis Anastasiadis-Katerini-Greece

+ + = ⇔ = − −

+ + ≥ ( + ) √ +

= + − ( + ) ≥ ( + ( + ) ) √ − ( + ) =

≤+

+ + ++

− ( + ) ⇒

≤ ( + + )( + )( − ( + ) ) It’s enough to prove:

+ − ( + ) ≥ ( + + )( + )( − ( + ) ) (1)

It is ( + ) − − = ( + )( + + ) (Cauchy)

(1) ⇔ − ( + )( + + ) ≥ ( + + )( + )( − ( + ) )

− ≥ ( − ( + ) ) ⇔ ( + ) ≥ which stands

5. If , , , ∈ ℝ then:

( − ) + ( + ) ≥ ( + ) ( + )


Solution 1 by Kevin Soto Palacios – Huarmey – Peru , Solution 2 by Serban George Florin-

Romania , Solution 3 by Fotini Kaldi-Greece , Solution 4 by Ravi Prakash-New Delhi-India ,

Solution 5 by Myagmarsuren Yadamsuren-Darkhan-Mongolia , Solution 6 by Soumava

Chakraborty-Kolkata-India

Solution 1 by Kevin Soto Palacios – Huarmey – Peru

Si , , , ∈ . Probar que

( − ) + ( + ) ≥ ( + ) ( + ) Ahora bien

www.ssmrmh.ro

( + )( + ) = ( + ) + ( − )

La desigualdad es equivalente

( − ) + ( + ) ≥ (( + ) + ( − ) )

⇔ (( + ) − ( − ) ) ≥

Solution 2 by Serban George Florin-Romania

( + )( + ) = ( + ) + ( − )

+ = , + = , − = , + =

⇒ = + , + ≥

= ( + ) ≤ ( + ) ⋅ ( + ) ⇒ ≤ +

Solution 3 by Fotini Kaldi-Greece

( + ) ≥ ( + ) , “ = “ if and only if =

( − ) + ( + ) ≥ (( − ) + ( + ) ) =

= ( + )( + )

Solution 4 by Ravi Prakash-New Delhi-India

Let = + , = + , then = ( − ) + ( + )

Now, ( + ) ( + ) = | | | | = (| |)

= [( − ) + ( + ) ] ≤ [( − ) + ( + ) ]

∵ ( + ) ≤ ( + )

Solution 5 by Myagmarsuren Yadamsuren-Darkhan-Mongolia

+ + + = + + +

: ( + − ) + ( + + ) =

= ( − ) + ( + ) = + + + =

www.ssmrmh.ro

= ( + )( + ) =

LHS: ( − ) + ( + ) = ( + )( + ): RHS ( ) = ( )

( − ) + ( − ) ⋅ ( + ) + ( + ) = = ( + ) ⋅ ( + ) ; ( − ) ≥ ; ( + ) > 0

( + ) ( + ) = ( − ) + ( − ) ( + ) + ( + ) ≤⏞ ≤ ( − ) + ( + )

Solution 6 by Soumava Chakraborty-Kolkata-India

( + ) + ( − ) = ( + )( + )

Given inequality ⇔ + ≥

⇔ ( + ) ≥

But, ( + ) ≥ ( + ) =

∵ + ≥ ( + )


If , ∈ [ , ] then:

++

++ ( − ) ≤


Solution 1 by Redwane El Mellas-Casablanca-Morocco

Solution 2 by Richdad Phuc-Hanoi-Vietnam

www.ssmrmh.ro


Let ( , ) = + + ( − ) .

∴ ( , ), ( > 0,0), ( , > 0) = , , + ≤

Suppose that , > 0:

If ≤ : ( , ) ≤ + + ( − ) = + ≤ ≤

If > 1: ∴ ( , ) − ( , ) = + + ( − ) − +

= + − − ( + )

=+

− − ( + ) ≤ − − ( + ) < 0

⇒ ( , ) < ( , ) = + ≤

Finally ( , ) ≤ with equality if and only if = , = .

Solution 2 by Richdad Phuc-Hanoi-Vietnam

We have + + ( − ) ≤ ⇔ − + − + − ≤

⇔( − )( + + ) −

+ + ( − )( + ) + + − ≤

⇔( − )( + + )

+ + ( − )( + ) + + − − + ≤

which is true, because − − ≤ − − ≤ (∀ , ∈ [ , ])

Equality holds if and only if = ; =

www.ssmrmh.ro

7. If < ≤ ; , , > 0

= , = , = then:

+ + ≥+ ++ +


Solution by Nirapada Pal-Jhargram-India

=∑

≥⏞∑

= ∑ =

− =∑

−∑∑ =

(∑ ) − ∑∑ =

∑ − ∑∑ ≥

− =∑∑ − ∑ =

(∑ ) − ∑∑ ∑ =

∑( ) − ∑( )( )∑ ∑ ≥

since + + ≥ + + , so ≥ ≥

And < ≤ ⇒ ≤ ≤

So, ≤⏞( )

= + +

8. If < ≤ ≤ , , ≥ then:

( + + )( + ) + ( + ) ≥

≥ ( + + )( + + )


Solution by Kevin Soto Palacios – Huarmey – Peru

Si < ≤ ≤ , , ∈ ℝ. Probar que

( + + )( + ) + ( + ) ≥

≥ ( + + )( + + )

www.ssmrmh.ro

Condición ≥ ≥ ⇔ − − = ( − ) + ( − ) ≥

Desarrollan do la desigualdad

+ + + + + + + ≥

≥ + + + + + +

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

⇔ − ( − ) − ( − ) + ( − ) = ( − ) ( − − ) ≥ (LQQD)

9. If ≤ , ≤ then:

++

++ ( − ) ≤ +


Solution by Chris Kyriazis-Greece

The function ( , ) = + + ( − ) − −

≤ ≤ , ≤ ≤ is convex due to or

(it’s easy to check it with positive derivatives)

So the function achieves its maximum to one of the vertices of the square

[ , ] × [ , ]; ( , ) = + + − − =

( , ) = + + − − = − − < 0

( , ) = + + − − = − − < 0

( , ) = + + − − = − < 0

So the maximum is zero when = and = .

www.ssmrmh.ro

10. If , , > 0 then:

( + + )( + + + + − ) + ≥ + +

Proposed by Rustem Zeynalov-Baku-Azerbaidian

Solution by proposer

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

≥ + + + ; + = ; + =

( + )( − + ) + ≥ + ; + + ≥ +

+ + ≥ =+ + ≥ =

+

+ + ≥ +

11. If , , > 0, 6 = then:

( + )( + )( + )≥ ( + + )


Solution 1 by Leonard Giugiuc – Romania, Solution 2 by Kevin Soto Palacios –

Huarmey – Peru, Solution 3 by Myagmarsuren Yadamsuren-Darkhan-

Mongolia

Solution 1 by Leonard Giugiuc – Romania

= , = , = ⇒ ( + + ) = ⇒

⇒ = , = , = ⇒ + + = and

=√

, =√

, =√

. After usual algebraic computations, we

get the equivalent + ∑ + ( + + ) ≥ ( )

www.ssmrmh.ro

By AM – GM, ≤ . So that

≥ ( ) , by AM – GM ∑ ≥ ( ) ; but

( ) ≥ ( ) ⇒ ≥ ( )

and by AM – QM + + ≥ ⇒ ( + + ) ≥ ( ) .


Siendo , , > 0, 6 = ( ). Probar que

( + )( + )( + )≥ ( + + ) =

Realizando los siguientes cambios de variables

= > 0, 2 = > 0, 3 = > 0, con , , > 0

La condición es equivalente

( + + ) = ⇔ + + =

⇔ + = + + + = ( + )( + )

⇔ + = ( + )( + ), + = ( + )( + )

La desigualdad propuesta es equivalente

( + )( + )( + ) ≥

( + )( + )( + )( + )( + )( + ) ≥

⇔ ( + ) ( + ) ( + ) ≥ ( )( )( ) =

(Válido por MA ≥ MG)


= ⋅ ( + + )

LHS: a) + ⋅ ( + + ) ⋅ + ⋅ ( + + ) ⋅

www.ssmrmh.ro

⋅ + ⋅ ( + + ) =

= ⋅ + ( + + ) ⋅ + ( + + ) ⋅

⋅ ( + + + ) =

= ⋅ ( + + + ) ⋅ ( + + + ) ⋅

⋅ ( + + + ) ≥⏞

≥ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ = ⋅ ⋅ =

= ⋅ ⋅ = a)

:≥⋅

= = ( + + )

12. If < < < then:

√ + √ + + √ − √ + < √ + √ +


Solution by Ravi Prakash-New Delhi-India

Let ( ) = + , ≥ , ( ) = − = − > 0,∀ > 2

Thus, increases on [ ,∞). For < < < , then

( ) < ( ) < ( ) < ( ). We have

√ + √ + + √ − √ +

= √ + √ ( ) + √ − √ ( ) < √ + √ ( ) + √ − √ ( )

www.ssmrmh.ro

= √ + √ ( ) < √ + √ ( ) = √ + √ +

13. If ≤ < < < then:

√ + √ + √ + √ > √ + √ + √


Solution by Myagmarsuren Yadamsuren-Darkhan-Mongolia

= √ + √ + √ + √ > √ + √ + √

:

√ > √

√ > √

√ > √

√ > √ ⎭⎪⎪⎬

⎪⎪⎫

(*)

> √ + √ + √ + √ (**)

√ = + + √√ = + + ⋅ √√ = + ⋅ √

√ = √ ⎭⎪⎪⎬

⎪⎪⎫

(***)

(**), (***) ⇒ > + + ⋅ √ + + + ⋅ √ +

+ + ⋅ √ + ⋅ √ =√

+√

+√

+

+√

+ ⋅ √ + √ + √ + √ + √ + √

www.ssmrmh.ro

≥ ⋅√ ⋅ √ ⋅ √

+ ⋅√

⋅√

⋅√

⋅√

⋅√

⋅√

+

+√

⋅√

⋅√

⋅√

⋅√

⋅√

⋅√

⋅√

⋅√

⋅√

=

= √ + √ + √ ⋅ ⋅ ⋅

14. If , , > 0, + + = then:

− + <


Solution 1 by Dat Vo-Quynh Luu-Vietnam,

Solution 2 by Khanh Hung Vu-Ho Chi Minh-Vietnam

Solution 1 by Dat Vo-Quynh Luu-Vietnam

− + =+ −

→ inequality ↔ + − < 3( + + )

↔ ( − ) − ( − ) + > 0 ↔ ( − + ) + ( − − ) > 0


If , , > 0, + + = . Prove that − + < (1) We have

(1)⇒ < ⇒ + + > 6 − +

⇒ + − + + − > 0 ⇒

⇒ + ( − ) + + − > 0 (2)

www.ssmrmh.ro

( ) = + ( − ) + + −

We have = ( − ) − ⋅ ( + − ) = − ( − ) < 0

⇒ (2) true ⇒ QED

15. If ≤ < 3, 0 ≤ < 5, 0 ≤ < 7 then:

√ + + √ + + √ + < 6


Solution 1 by Nirapada Pal-Jhargram-India, Solution 2 by Seyran Ibrahimov-

Maasilli-Azerbaidian, Solution 3 by Nguyen Thanh Nho-Tra Vinh-Vietnam,

Solution 4 by Nguyen Ngoc Tu-Ha Giang-Vietnam

Solution 1 by Nirapada Pal-Jhargram-India

√ + + √ + + √ + < 6

≤⏞ + + = + +

< + + . As ≤ < 3, 0 ≤ < 5, 0 ≤ < 7

= + + =

Solution 2 by Seyran Ibrahimov-Maasilli-Azerbaidian

1) √ + < 2 ⇒ < 7 ⇒ 0 ≤ < 3 True

2) √ + < 2 ⇒ < 31 ⇒ 0 ≤ < 5 True

3) √ + < 2 ⇒ < 127 ⇒ 0 ≤ < 7 True

(1)+(2)+(3) < 6 (proved)

Solution 3 by Nguyen Thanh Nho-Tra Vinh-Vietnam

< 3 ⇒ + 1 < 4 < 8 ⇒ √ + < √ =

www.ssmrmh.ro

< 5 ⇒ + 1 < 6 < 32 ⇒ √ + < √ =

< 7 ⇒ + 1 < 8 < 128 ⇒ √ + < √ =

√ + + √ + + √ + < 6

Solution 4 by Nguyen Ngoc Tu-Ha Giang-Vietnam

We have √ + < √ + < 2√ + < √ + < 2√ + < √ + < 2

⇒ √ + + √ + + √ + < 6,

∀ ≤ < 3, 0 ≤ < 5, 0 ≤ < 7

16. If < ≤ then:

≤++

+++

+++

≤


Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by Ravi

Prakash-New Delhi-India, Solution 3 by Soumitra Mandal-Chandar Nagore-

India, Solution 4 by Abdallah El Farissi-Bechar-Algerie, Solution 5 by SK

Rejuan-West Bengal-India, Solution 6 by Antonis Anastasiadis-Katerini-Greece


Siendo < ≤ . Probar que ≤ + + ≤

Como ≥ > 0

⇒ = ⋅ ≥ , = ⋅ ≥ , = ⋅ ≥

⇒ = ⋅ ≤ , = ⋅ ≤ , = ⋅ ≤

Por lo tanto

⇒ + + ≥ + + = + + =

www.ssmrmh.ro

⇒ + + ≤ + + = + + =


For < ≤ , ∈ ℕ, + ≤ +

≤ + ≤ + = + ⇒ ≤++ ≤

Taking = , , and adding we get the desired inequality.


++

+++

+++

≥

⇔++ − +

++ − +

++ − ≥

⇔ ( ) + ( ) + ( ) ≥ , which is true ∵ ≥

≥++

+++

+++

⇔ ( ) + ( ) + ( ) ≥ , which is true ∵ ≥

∴ ≤++

+++

+++

≤

Solution 4 by Abdallah El Farissi-Bechar-Algerie

Let , ∈ ℝ∗, ≤ for all ∈ [ , ], ≤ + ( − ) ≤

([ , ] is a covex set)

We have = + = + −

Then for ≥ , ≤ ≤ it follow that ≤ ∑ ≤

Solution 5 by SK Rejuan-West Bengal-India

www.ssmrmh.ro

Applying Cauchy inequality we get,

≥ ≥⇑

[ ]

(1)

Again applying Cauchy inequality we get,

≥ ≥⏞ ≥ (i2)

[from (i)]. Again applying Cauchy inequality we get,

≥ ≥⏞ ≥ (iii)

[from (ii)]. Adding (1), (ii) & (iii) we get,

+ + ≥ ( + ) ≥ [Proved]

[as ≥ ⇒ + ≥ ]. Applying Cauchy inequality we get,

≤ ≤⏞ (1)

≤⏞ ≤⏞ (2)

From (1) & (2) we get, ≤ ; ≤

∴++

+++

+++

≤++

[ ≤ ] ⇒≤+ ⋅+

= ⋅++

=

⇒ + + ≤ [proved]

Solution 6 by Antonis Anastasiadis-Katerini-Greece

www.ssmrmh.ro

17. If , , ≥ then:

+ + ≥ √ + √ √


Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by Seyran

Ibrahimov-Maasilli-Azerbaidian, Solution 3 by Abdallah El Farissi-Bechar-

Algerie


Siendo , , ≥ . Probar que + + ≥ √ + √ √

Como , , ≥ . Aplicando ≥

+ + + ≥ √ (A)

+ + + ≥ √ (B)

+ + + ≥ √ (C)

Sumando (A) + (B) + (C) ( + + ) ≥ √ + √ + √ √ ⇔

⇔ + + ≥ √ + √ + √ √ ≥ √ + √ √


= , = , = , + + ≥ ( + )

+ + ≥ ( + + )( + + ) ≥ ( + + )

www.ssmrmh.ro

( + + ) ≥ ( + ) ⇒ ≥


√ + √ √ = √ √ + √ √ ≤+ √ + + √

≤

≤+ + + + +

=+ +

≤ + +

18. If ≤ < < then:

−+

−+

−<

−+

−+

−


Solution 1 by SK Abdul Halim-India, Solution 2 by Ravi Prakash-New Delhi-

India

Solution 1 by SK Abdul Halim-India

If , be a positive real number but is not equal to and , are rational

numbers then: > if >

Proof: > 0 and ≠ . > 0 and ≠

If = then both sides are equal with zero.

Let two unequal numbers and with associated positive rational

weights and − respctively, then we know. ⋅ ( )⋅ > [ − ] or, + − >

or, ( − ) > ( − ) or, > , since > 0, > 0

Now, for = , = ; > (i) ≤ < <

www.ssmrmh.ro

Similarly, > (ii)

From (i) – (ii) we get, − > −

i.R. >

Rest are in similar member. Hence the proof.


Let ∈ ℕ and > 1. Consider ( ) = − , ∈ [ , ],

( ) = − , ∈ [ , ]

By the Cauchy’s mean value theorem ∃ ∈ ( , ) such that

( ) − ( )( ) − ( ) =

( )( ) ⇒

−−

=( + )

=+

>+

; [∴ > 1] ⇒−

+>

−

If > ≥ 1, then

−+

>−

⇒−+

>−

≥−

Thus, > ,∀ > ≥ 1, ∈ ∴ if ≤ < < , then

−+

−+

−<

−+

−+

−

19. If < ≤ ≤ then:

( + )√ + ( + + )√ +( + + ) + √ + √

≥


www.ssmrmh.ro

Solution 1 by Ravi Prakash-New Delhi-India, Solution 2 by Geanina Tudose-

Romania


Let < ≤ ≤ 1. Consider ( + ) + ( + + )( ) + −

−( + + ) + + ( )

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

− − − − ( + ) − − ( + + )( ) − ( )

= ( + − ) + ( − − ) + ( + − )( )

= ( − − ) − − ( ) ≥

as < , ≤ , < ; ( ) ≤ ; ( ) ( )( )

( ) ( )≥

Put = , = , we get ( )√ ( )( )

( ) √ ( )≥

Solution 2 by Geanina Tudose-Romania

( + )√ + ( + + )√ + ≥

≥ ( + + ) + √ + √

⇔ ( + )√ + ( + + )√ + ≥

≥ ( + + ) + ( + )√

+ √ + ( + + )√ + + √ + √

⇔ ( + )√ + ( + + )√ + ≥ ( + + ) +

+ √ + √

www.ssmrmh.ro

⇔ ( + − )√ + ( + − )√ + ( − − ) ≥

⇔ ( − − ) − √ − √ ≥ (1)

AM ≥ GM we have √ ≤ ⋅ = ( + ) ≤

√ ≤ ≤ = . Hence − √ − √ ≥

and − − ≥ . Therefore (1) is true

20. If , , > 0, = 1 then:

+ + ( + )( + ) + ( + )( + )( + ) ≥


Solution 1 by Nguyen Minh Tri-Ho Chi Minh-Vietnam, Solution 2 by Abdallah

El Farissi-Bechar-Algerie, Solution 3 by Ravi Prakash-New Delhi-India, Solution

4 by Tuk Zaya-Ulaanbaatar-Mongolia, Solution 5 by Nguyen Thanh Nho-Tra

Vinh-Vietnam, Solution 6 by Soumitra Mandal-Chandar Nagore-India

Solution 1 by Nguyen Minh Tri-Ho Chi Minh-Vietnam

( + )( + ) + ( + ) + ≥ ( + )( + )( + )

⇔ ( + + + ) + + + ≥ ( + )( + + + )

⇔ + + + + + + ≥ ( + + + + + + + )

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

⇔ ( + + ) + ( + + ) ≥ (1)

We have: + + ≥ √ = ;

+ + ≥ = ⇒ (1) true⇒ Q.E.D


If = , then + ( )( ) + ( )( )( ) ≥

www.ssmrmh.ro

We have ( + )( + )( + ) ≥ √ = then − ( )( )( ) ≥

and + ( )( ) + ( )( )( ) = − ( )( )( )

generalization if ≥ ( = , , … , )−∏ = , then

∏ ( + ) ≥−


As = , = , = + ( )( ) + ( )( )( )

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

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

( + )( + )( + )

=+ + + ( ) + + +

+ + + + + + + ( )

= − ( )( )( ). But + ≥ √ , + ≥ √ , + ≥ √

⇒ ( + )( + )( + ) ≥ ⇒ ≥ ( + )( + )( + )

Thus, ≥ − =

Solution 4 by Tuk Zaya-Ulaanbaatar-Mongolia

, , > 0 = 1; +

+ ( + )( + ) + ( + )( + )( + ) ≥

+ = ⇒ = − ; + = ⇒ = − ; + = ⇒ = −

−+

−+

−≥ ⇔ − + − + − ≥

www.ssmrmh.ro

≥ ⇔ ≥ ⇔ ( + )( + )( + ) ≥

+ ≥ √+ ≥ √+ ≥ √

⇔ ( + )( + )( + ) ≥ √ = ; = = =


, , > 0, = 1; ( + )( + )( + ) ≥ + √ =

⇔ ( + )( + )( + ) ≤ ⇔ − ( + )( + )( + ) ≥ −

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

⇔+

+ ( + )( + ) + ( + )( + )( + ) ≥


++ ( + )( + ) + ( + )( + )( + )

= − + + ( + )( + ) + ( + )( + )( + )

= − ( )( ) + ( )( )( ) = − ( )( )( )

≥⏞ − = − =

21. If , , , ≥ then:

( + ) ( + ) ( + )( + )( + )( + ) ≤ ( + )( + )( + )


www.ssmrmh.ro

Solution 1 by Abdul Aziz-Semarang-Indonesia, Solution 2 by Geanina Tudose-

Romania

Solution 1 by Abdul Aziz-Semarang-Indonesia

Since , , ≥ , we have

)( − )( − ) ≥ ⇒ ( + )( + ) ≤ ( + ) (1)

)( − )( − ) ≥ ⇒ ( + )( + )( + ) ≤ ( + ) (2)

)( − )( − ) ≥ ⇒

( + )( + )( + )( + ) ≤ ( + )(3)

Multiplying (1), (2), (3)

( + ) ( + ) ( + ) ( + ) ≤ ( + )( + )( + )

⇔( + ) ( + ) ( + )

( + )( + )( + ) ≤ ( + )( + )( + )

Solution 2 by Geanina Tudose-Romania

Rewritting the inequality

( + ) ⋅ ( + ) ⋅ ( + ) ⋅ ( + ) ≤ ( + )( + )( + )

We show ( + )( + ) ≤ ( + ) (1)

( + )( + )( + ) ≤ ( + ) (2)

and ( + )( + )( + )( + ) ≤ ( + ) (3)

In fact, we have more generally

( + )( + ) … ( + ) ≤ ( + … )

for (∀) ≥ , ≥ , > we show it by induction:

( ): ( + )( + ) ≤ ( + )

⇔ + + ≤ + ⇔ ( − )( − ) ≥ true

( ) → ( + )

www.ssmrmh.ro

( + ) … ( + )( + ) ≤⏞( )

( + … )( + ) ≤⏞( )

≤ ⋅ ⋅ ( + … )

Thus multiplying (1), (2), (3) we have the desired inequality

22. If < ≤ ≤ < 1 then:

( − ) + ( − ) + ( − ) < −


Solution by Abdelhak Maoukouf-Casablanca-Morocco

< ≤ ≤ < 1

= ( − ) + ( − ) + ( − )

= ( − ) + ( − + − ) + ( − )

= ( − )( + ) + ( − )( + )

= ( − )+− + ( − )

+−

“ ≤ < 1”; “ < 1; < 1”

< ( − ) + ( − ) = ( − ) < < −

23. If , , , ≥ then:

+ + + + + + + >+ √

++ √

++ √



Siendo , , , ≥ . Probar que

www.ssmrmh.ro

+ + + + + + + >+ √

++ √

++ √

Es suficiente probar + ≥√

(A)

⇔+

− ++

≥+ √

− ⇔+

−+

≥− √+ √

⇔

⇔−

( + )( + ) ≥− √+ √

⇔ √ √( )( ) ≥ √

√⇔ −√ ( √

( )( ) − √=

= − √+ √ − ( + )( + )

( + )( + ) + √=

= √ √ √( )( ) √

≥ , lo cual es cierto ya que → ≥

Analogamente para los siguientes términos

+ + +√

≥√

+⋅ √

≥√

⇔

⇔ + + ≥√

(B)

⇒ + + + ≥√

+√

≥√

(C)

Sumando (A)+(B)+(C)

⇒+

++

++

++

≥+ √

++ √

++ √

Por lo tanto

+ + + = + + + + + ≥

www.ssmrmh.ro

≥+ √

++ √

++ √

+ + + + >

>+ √

++ √

++ √

24. If < ≤ ≤ then:

( − ) ⋅ ( ) + ( − ) ⋅ ≤ ( − ) ⋅ ( ) + ( − ) ⋅



∀ < ≤ ′ ( ) = ( − ) − ( − )

⇒ ( ) =−

−( − )

=− ( − ) ( − )

( − ) > 0

≤ ⇔ ( ) ≤ ( ) ⇔ ( ) ≤

< ≤ ≤ ⇒ ( ) ≤ & ( ) ≤ & ( ) ≤

⇒ ( ) + ( ) + ( ) ≤

⇔ [ ( − ) − ( − ) ] + [ ( − ) − ( − ) ] +

+[ ( − ) − ( − ) ] ≤

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

25. If , > 0, + + = 0 then:

+ + ≥ ( + ) √ +


Solution by Anas Adlany-El Zemamra-Morocco

First, note that: ( + + ) = + + + ( + )( + )( + )( + + + + + )

www.ssmrmh.ro

Then when + + = , we will have the following

+ + = ( + + + + + );

Now, let’s go back to the main problem and check out what we are really

dealing with the problem asks us to show that:

+ + ≥ ( + ) √ + whenever , > 0

We have + + ≥ ( + ) √ +

⇔ ( + + + + + ) ≥ ( + ) √ +

⇔ ( + + )( + + + + + ) ≥ ( + ) √ +

and the last step can be explained as follows:

+ + = ⇒ ( + ) + = ⇒ + + + ( + ) =

⇒ + + = ( + = − )

Now, since , > 0; by the AM-GM inequality we get:

+ + ≥ √ + − √ + (1)

Also, ∑( + ) − − = + + + +

= ( + ) + + ( + )( + = − )

= (( + ) − ) − = − − = − = ( − ) ≥

Which prove that ∑( + ) ≥ + (2)

Finaly, from results (1) & (2) the proof is completed.

26. If < ≤ ≤ then:

( − )( − )( − ) √ √ √ ≤ − ( − ) −


www.ssmrmh.ro

Solution 1 by Chris Kyriazis-Greece, Solution 2 by Abdelhak Maoukouf-

Casablanca-Morocco, Solution 3 by Ravi Prakash-New Delhi-India, Solution 4

by Soumitra Mandal-Chandar Nagore-India

Solution 1 by Chris Kyriazis-Greece

If = or = or = , the inequality is obvious

Let’s suppose that < <

The function ( ) = is strictly convex, so using the Hermite-

Hadamard inequality we have that: ∫ ( )

≥ ( ) ⇔ ≥

⇒⏞√

− ≥ ( − ) √ . Doing exactly the same:

− ≥ ( − ) √ ; − ≥ ( − ) √

Multiplying those three inequalities (everything is positive) we have

that − ( − ) − ≥ ( − )( − )( − ) √ √ √

Solution 2 by Abdelhak Maoukouf-Casablanca-Morocco

∵ ∀ ≥ ( ) ≥ So ∀ ≥ : ≥

⇔ ≤ ⇔ ( − ) ≤ − (1)

∵ ∀ , > 0: ≤ ⇔ ≤ (2)

(1)×(2) ⇒ ∀ ≥ > 0( − ) ≤ −

⇔ ( − ) ≤ ( − )

www.ssmrmh.ro

< ≤ ≤ ⇒( − ) √ ≤ −( − ) √ ≤ −( − ) √ ≤ ( − )

⇒ ( − )( − )( − ) √ √ √ ≤ − ( − ) −


If = , − = ( − ) √ . Suppose > > 0

−−

=−

( − ) +!

( − ) +!

( − ) + ⋯

= + !( + ) + ! ⋅

( + + )+ ! ⋅

( + + + ) + ⋯

> 1 + √ + !( ) + !

( ) + ⋯

= + √ + !√ + !

√ + ⋯ = √

⇒ − > √ ( − )∀ ≥

Thus, ( − ) √ ( − ) √ ( − ) √ ≤ − ( − ) −

⇒ ( − )( − )( − ) √ √ √ ≤ − ( − ) −


is convex function. Hence, applying Hermite – Hadamard – Inequality

∫ ≥ , ∫ ≥ and ∫ ≥

⇒ − − ( − ) ≥ ( − )( − )( − )

≥ ( − )( − )( − ) √ √ √

27. If < < < < then:

www.ssmrmh.ro

>


Solution 1 by Abdelhak Maoukouf-Casablanca-Morocco, Solution 2 by

Abdallah Almalih-Damascus-Syria


∴ let ( ; ) = ; ( ; ) ∈ (ℝ∗ )

⇒ ( , ) =( − ) − ( − )

( − ) =− +

( − ) ≤

∵ + ≤ :∀ > 0 ∗ ( ; ) = ( ; ) ⇒ ( ; ) ≤

∵ < << < ⇒ ( ; ) > ( ; ) ⇔

−− >

−−

⇔ > ⇔ >

Solution 2 by Abdallah Almalih-Damascus-Syria

If < < < < then

> this inequality equivalent to ( is increased function)

> ; > which it’s right.

because ( ) = is a concave function as ( ) = − < 0 So it satisfy

( ) ( ) ≤ ( ) ( ) ≤ ( ) ( ) for ∈] , [ take = ∈] , [

so ( ) ( ) < ( ) ( ) < ( ) ( ) (1)

www.ssmrmh.ro

take ∈ ∈] , [ so ( ) ( ) < ( ) ( ) < ( ) ( ) (2)

hence ( ) ( ) <( ) ( ) ( ) <

( ) ( ) ( ) so ( ) ( ) < ( ) ( )we have

( ) = so ( ) ( ) ≤ ; ≤

28. If ≤ , , < 1 then:

( + )( + )( + )( − )( − )( − ) ≥

+−



For ≤ , , < 1, consider = ( − ) ( + )( + )( + ) −

−( + ) ( − )( − )( − )

=( − ) ( + )

( − )( − )( − ) ( + )( + )( + )

Use → − to obtain

=( − ) +

( − )( − )( − ) ( + + ) +

Use → +

= + ++ + + + + +

Use → −

= + ++ + − + + −

Note that + − ( + ) = ( − ) > 0

⇒ + > 3 + (1)

www.ssmrmh.ro

Also, + + ≥ + + and ( − ) + ( − ) + ( − ) ≥

≥ ( − ) + ( − ) + ( − ) [∵ ≤ , , < 1]

⇒ ( + + )[( − ) + ( − ) + ( − ) ]

≥ ( + + )[ ( − ) + ( − ) + ( − ) ]

⇒ + + − ≥ + + − (2)

From (1), (2), we get ( + )( + + − ) ≥

≥ [ + + − ] ⇒ ≥

⇒ ( − ) ( + )( + )( + ) ≥ ( + ) ( − )( − )( − )

Put = , = , = to obtain

( + )( + )( + )( − )( − )( − ) ≥

( + )( − )

29. If < < < then:

−−

+>

−−

+


Solution 1 by Abdelhak Maoukouf-Casablanca-Morocco, Solution 2 by

Soumitra Mandal-Chandar Nagore-India


∴ let ( ; ) = ; ( ; ) ∈ (ℝ∗ )

⇒ ( , ) = + ( − ) − ( − )

( − )

www.ssmrmh.ro

= −( )

( ) ≤ ∵ ≤ ≤ :∀ ≥

∗ ( ; ) = ( ; ) ⇒ ( ; ) ≤

∵ < << < ⇒ ( ; ) > ( ; ) ⇔ >

∵

⎝

⎛

+

=+−

∀ < 1 ⎠

⎞ ⇔−

+− >

−+

−


LEMMA: For any concave function : [ , ] → ℝ and ∈ ( , ) we have ( ) − ( )

− ≤( ) − ( )

− ≤( )− ( )

−

Let ( ) = for any ∈ [ , ], ( ) = , ( ) = − ( ) ≤

so, is concave, hence for < < we have

< < (1)

for < < , < < (2)

combining (1) and (2) we have, <

(Proved)

30. If < ≤ ≤ , + + = 3 then:

+ + ≥ + + + ( − )( − )( − )


Solution 1 by Abdul Aziz-Semarang-Indonesia, Solution 2 by Pham Quoc Sang-

Ho Chi Minh-Vietnam


www.ssmrmh.ro

+ + + ( − )( − )( − ) = + ( )( )( )

+ + = ⇒ ≤⇓

≤+ + + − + − + −

=+ +

= + +

Solution 2 by Pham Quoc Sang-Ho Chi Minh-Vietnam

≥ + ( − )( − )( − )

Because + + = ⇒ ≤ =

We have: ∑ − ∑ = − + − + −

= [ ( − ) + ( − ) + ( − )]

= [ ( − ) + ( − + − ) + ( − )]

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

≥ ( − )( − )( − )

(because ≤ and − ≥ , − ≥ , − ≥ )

= ( − )( − )( − ); “=” = = =

31. Let , , be positive real numbers. Prove that

+ + ≥ ( + + ) + ( )( )( )( )( )

Solution by Nguyen Ngoc Tu – HaGiang – Vietnam

www.ssmrmh.ro

Solution by proposer

We have ∑ = ∑ ( )( ) ( )

=( − )( − )

++

( − )( − )+

+ ( + + )+

=+

( + )( + )( − ) + ( + + )

( )( )( − ) = ( )

( )( )( )( − ) ≥ ( )( )( )

( − )

= ( + )( + )( + ) ⋅( − )

Similarly ( )( )( − ) ≥ ( )( )( ) ⋅

( ) ,

+( + )( + ) ( − ) ≥ ( + )( + )( + ) ⋅

( − ).

Hence ∑ ( )( )( − ) ≥ ( )( )( ) ⋅

( )

=( − )

( + + )( + )( + )( + )

⇒ + + ≥ ( + + ) + ( )( )( )( )( )

32. If , , ≥ , + + = then:

( + + ) + ≥ ( + + )

Proposed by Sladjan Stankovic-Macedonia

Solution 1 by Soumitra Mandal-Chandar Nagore-India, Solution 2 by Le Khanh

Sy-Long An-Vietnam, Solution 3 by Marian Cucoanes – Romania, Solution 4

www.ssmrmh.ro

by Leonard Giugiuc – Romania, Solution 5 by Ngo Minh Ngoc Bao-Vietnam ,

Solution 6 by Marian Dincă – Romania


Let , , ≥ , + + = then ∑ + ≥ ∑

We will prove that for any , , ≥ with + + = the inequality

( ) ≥+ +

… . ( )

holds true where ( ) = − + for all ≥

Now, ( ) = − > 0 for all ≥ = = .

So, the function is right – convex for ≥ . By RCF theorem it suffices

to prove (1) for = ≥ ≥ with + + = . Hence we will show

( ) ≥ ( ) for ≥ ≥ ≥ and + =

where ( ) = ( ) ( ) = + − − . So, ( ) − ( )

= ( − ){ ( + + ) + ( + )− } = ( − )( − + )

[∵ + = ] = ( − )( − )( − ) ≥ [∵ + = ⇒ ≥ − > 0]

∴ ( ) ≥ ( ) for ≥ ≥ ≥ and hence the proof is complete. So,

+ ≥

(Proved) Equality at = = = and = = , = or any other

permutations.

Solution 2 by Le Khanh Sy-Long An-Vietnam

Let , , be nonnegative real numbers such that + + = and

www.ssmrmh.ro

= { , , }. Prove that

( + + ) + ≥ ( + + ) +( − )

Lemma: If , , ≥ and = { , , } → + ≥ , then

( + ) − ( + ) ≥( + )

−( + )

+( − )

Or ( )

≥ ( − )

We have =( ) ( )

≥

Thus, the proof is completed. Therefore, it suffices to ( ) + + ≥ + ( ) , Which is ( − ) ( + ) ≥

The equality occurs for = = = , and for = and = = .

Solution 3 by Marian Cucoanes – Romania

+ + = (1); ( + + ) + ≥ ( + + ) (2)

(2) ⇔( )

( + + ) + ( + + ) ≥

≥ ( + + ) ( + + ) and because:

( + + ) + ( + + ) − ( + + ) ( + + ) =

= ( + + + + + )− ( + +

+ ) − ( + + + + + ) ⇔

⇔ ( + + + + + ) ≥ ( +

+ + ) + ( + + + + + ) (3)

Using Schur inequality for degree = ⇒

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

⇔ + + + + + ≥ +

www.ssmrmh.ro

+ + + + + ⇒

( + + + + + ) ≥ ( +

+ + + + + ) (4)

Using AM – GM inequality ⇒ ( + + + + + ) ≥

≥ ( + + ) (5)

Then (4) + (5) ⇒ (3) ⇒ (2) ⇒ q.e.d.


It’s well known that,

( + + ) + ( + + ) ≥ ( + + )( + + ) ⇒

( + + ) ≥ ( + + ) − . Suffice it to show that

( + + ) − + ≥ ( + + ). But

+ + = ( + ) ⇒ + + = ( + ), where = .

The latter is + − ≥ . But if ≤ ≤ , then

≥ − − + , hence suffice it to show − ≥ , which is true.

If ≤ ≤ , then + − ≥ − ≥ .

Done! Equality at ( , , ) or , , and permutations.

Solution 5 by Ngo Minh Ngoc Bao-Vietnam

( + + ) + ≥ ( + + ) (*). We have lemma:

( , , ) = + + + + .

( , , ) ≥ ,∀ , , ≥ ⇔

⇔

+ + + + ≥( + ) < + +

+ + + ≥( ) = ( + + )( + ) + ( + − − ) + ( − + − ) ≥ ,∀ ≥

www.ssmrmh.ro

The inequality (*) ⇔

⇔ ( + + ) + ( + + ) ≥ ( + + ) ( + + )

⇔ + + + + + ≥

≥ + + + +

⇔ − − − + ≥ ⇔

⇔ − − − + ≥

Consider quadratic symmetry polynomial:

( , , ) = − + − −

Use lemma with = − , = , = = − ⇒

⇒

+ + + + =( + ) = < + + =

+ + + = > 0 (1)

Consider function: ( ) = − − + ,∀ ≥ .

We have: ( ) = ( − − ), ( ) = ⇔ − − = ⇔

⇔=

= −

www.ssmrmh.ro

⇒ ( ) ≥ ( ) = > 0,∀ ≥ 0 (2)

With (1) and (2) ⇒ ( , , ) ≥ .

Equality when = = = or ; ; or ; ; or ; ;


= − , = + , = ; + = −

+ = ( + ) − ( ) = ( − ) −

( + + ) + ≥ ( + + ) ⇔

[( − ) − + ( − ) ] + ≥ ( − + ( − ) )

Let : , →

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

( ) = (− )( ) ( − ) − + =

= − + − + = − ≤ ⋅ − =−

< 0

= decreasing ⇒

( ) ≥ = − − + ( − ) + − − + ( − ) =

= − + ( − ) + − + ( − ) =

= + ( − ) + − ⋅ − ( − ) = ( )

( ) = − ( − ) − + ( − ) =

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

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

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

www.ssmrmh.ro

− + = ⇒ , =± √ − ⋅

=± √

=

= ±√

, for = { , , }, result ≥ , + ≤ ⇒ ≤ and < < , we

obtain: − + = ( − )( − ) > 0

and: ( ) = ( − )( − + ) ≤

= decreasing ( ) ≥ ( ) = + + − ⋅ − − =

= − = , we shall obtain: ( ) ≥ ( ) ≥

33. If ∈ ℕ, ≥ , , ≥ then:

+ + + + ⋯+ + ≤ ( − ) +

Proposed by Daniel Sitaru - Romania

Solution 1 by Soumava Chakraborty-Kolkata-India ,


Solution 3 by Uche Eliezer Okeke-Anambra-Nigeria


+ ≤ + ⇔ ( + ) ≤ ( + ) (∵ , ≥ )

⇔ ≤ ( + ) ⇔ { ( + ) − } ≥

⇔ { ( − ) + ( + ) } ≥ → true

∴ + ≤ + (1)

Again, + ≤ + ⇔ + ≤ ( + )

⇔ ≥ → true ∵ + ≤ + (2)

(1), (2) ⇒ + ≤ + is true, for = , =

Let us assume + ≤ + holds true

for = (some integer ≥ ); we shall prove.

www.ssmrmh.ro

Then show that + ≤ + will hold true for = + as

well + ≤ +

⇔ + ≤ ( + ) (a)

By our assumption, + ≤ +

⇒ + ≤ ( + ) (3)

Now, ( + ) = ( + )( + )

≥ ( + ) + (by our assumption and by using (3))

= ( + ) + +

= + + + + + ≥?

+

⇔ + + ( + − ) ≥

⇔ + + ( + ) + ( − ) ≥ → true,

∵ , ≥ ⇒ a is true

So, whenever + ≤ + is true for = ( ≥ , ∈ ℕ), then,

+ ≤ + is true for = + as well.

Hence, by the principle of mathematical induction,

+ ≤ + (b) (∀) ≥ , ∈ ℕ

(b), (1) ⇒ + ≤ + ∀ ≥ , ∈ ℕ

∴ + ≤ ( − ) +


Let ≥ , ≥ . Suppose = , ≥

www.ssmrmh.ro

= ≥ ⇒ = ≥ =

⇒ ( + ) = + + ≥ + + = ( + )

Suppose = + , ≥ ( ) ≥ ≥ , [∵ ≥ ]. Now,

( + ) = + + ≥ + +

∴ ( + ) ≥ ( + ) ∀ ≥ , ≥ Suppose ≥ > 0 then

+ ≥ + ⇒ ( + ) ≥ ( + )

⇒ ( + ) ≤ + ⇒ ( + ) ≤ ( − ) +

Solution 3 by Eliezer Okeke-Port Anambra-Nigeria

If ∈ ℕ, ≥ , , ≥ . The ∑ + ≤ ( − ) +

Equality case follows if and only if ⇒ = = ⇒ ≠ ≠ , , ∈ ℝ

< . It suffices to prove

+ < + (1)

WLOG: Assume <

(*) + = + ⇒ + = + +

(**) + = + +

Clearly since < ∀ ≥ ⇒ <

www.ssmrmh.ro

Conclusion + < + (True)

+ ≤ + = ( − ) +


If , ∈ ( ,∞) then:

++

++

+≥


Solutions 1,2 by Fotini Kaldi-Greece

Solutions 3,4 by Myagmarsuren Yadamsuren-Darkhan-Mongolia


> 0, > 0

+ +≥

+ +⇒ + + ≥

≥ ( + + ) , , , > 0

≥+

++

+⋅ +

⋅=

≥⋅

+ + +⋅

≥

≥⋅

( + ) =


Weighted Means > 0, > 0

www.ssmrmh.ro

≥+

⋅+

⋅⋅ +

⋅=

= ( + )( + ) +⋅

≥

≥ √ √ ⋅= = ⇒ ≥


ln ⋅ ln + 12 ln ⋅ ln

=+ ⋅

( ) = ⇒ ( ) ≥ CONCAVE

++

+ ⋅ ++

≥

≥⋅

≥∙ √ ⋅ ⋅ =


==

= ∈ ℕ⇒

++

++

+≥

www.ssmrmh.ro

≥√

+

⎝

⎜⎛

⋅

⎠

⎟⎞

+ = √ + + ≥

( − )y ≥ ⋅ √ ⋅ ⋅ ( ) = .

35. If , , ∈ ( ,∞) then:

+ + ≥( + ) ( + ) ( + )


Solution 1 by Nirapada Pal-Jhargram-India , Solution 2 by Fotini Kaldi-Greece ,

Solution 3 by Eliezer Okeke-Anambra-Nigeria

Solution 1 by Nirapada Pal-India

= =

≥⏞+ + +

=( + ) ( + ) ( + )


+ + =

= + + + + + + + + ≥⏞

≥+ + +

⇒

www.ssmrmh.ro

( + + ) ≥+ + +

⇒

⇒ ( + + ) ≥ ( ) ( ) ( ) ; “ =” ⇔ = =

Solution 3 by Eliezer Okeke-Port Anambra-Nigeria

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

⇒ ( + + ) + ( + + ) =+ +

+ ( + )

=∑

+ ( + )

= + + ++

++

++

≥⏞

⋅ = ( ) ( ) ( )

36. If , , ∈ , then:

+ + + + + >

Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania

Solution 1 by Nirapada Pal-Jhargram-India , Solution 2 by Seyran Ibrahimov-

Maasilli-Azerbaidian , Solution 3 by Soumava Chakraborty-Kolkata-India


For , , ∈ , , < 1, < 1, < 1

So, + +

www.ssmrmh.ro

= + ⋅ + + ⋅ + + ⋅

>+

++

++

> 1, .

≥ by Nesbbit. ∴ + + >


≥( + + )

( + + ) >

( + + ) > 3( + + )

Jensen: + + ≥

> 3( + + ); + + < 9


∑ > ∑ ≥ (Nesbitt)

∵ > , < < 0 < , , <

37. Let ( , , ) is continuous and symmetric function with

, , > 0. If ( , , ) ≥ √ ,√ ,

Prove that: ( , , ) ≥ √ , √ , √

Proposed by Sladjan Stankovic-Skopje

Solution by Marian Dincă – Romania

Let: = , = , = ; ( , , ) = ( , , ) ≥ , ,

Let: ( , , ) = ( , , ) − continuous and symmetric equivalent to:

www.ssmrmh.ro

If: : ( , , ) ≥ , , . Prove that:

: ( , , ) ≥+ +

,+ +

,+ +

Let ≥ , let ∈ ,

→( − , + , ) =

+,

+, ≤ ( , , )

result to ( , ) such that: ∈ ( , )

( − , + , ) ≤ ( , , ), for ∈ ( , )

let: − = ⇒ = −

and: ( , , ) ≥ , + − ,

and: , + − , = + − , ,

if: + − ≥ , result:

+ − , , ≥ + − − , + ,

Let: + − − = ⇒ = + − ( + + )

We obtain: + − , , ≥ , , , or

( , , ) ≥+ +

,+ +

,+ +

⇔

( , , ) ≥ √ , √ , √

38. If , , > 0, + + = √ then:

++ ≤


www.ssmrmh.ro



Solution 3 by Sanong Hauyrerai-Nakon Pathom-Thailand


For ≤ ≤ √ , let ( ) = √ +

Note is continuous on ,√ and

( ) = −+ √ −

++

, < < √

=+ √ − − −

+ √ − ( + )=

√ − − √ − +

+ √ − ( + )

( ) > 0 if < <√

= if =√

< 0 if √

< < √

∴ attains its maximum value at =√

and

√=

√+

√=

Now, + ( ) = √ + ( ) ≤

Similarly, for other two expressions.


We have, ( ) = − ( ) ≤ for all ≥

So, is concave. Applying Jensen +

≤+ +

=

www.ssmrmh.ro

similarly, ∑ ≤ so, ∑ + ≤

Solution 3 by Sanong Haueyrerai-Nakon Pathom-Thailand

We will show that

+ + + + + ≤

Because = ,− < < , ∈ ℝ

If = ,− < < , ∈ ℝ and since + + = √ , , , > 0.

Hence = , < < , ∈ ℝ . If = , < < , ∈ ℝ

Give = , = , =

If = , = , =

and + + = + + = √

Hence + + = + + ≤ −

If = , = , =

and + + = + + = + + = √

Hence + + = + + ≤ =

Hence + + ≤ −

Therefore + + + +

+ ≤ + = . That is, it is to be true.

39. Prove that for any real numbers , ,

( + + )( + − )( + − )( + − ) ≤

Proposed by Nguyen Viet Hung – Hanoi – Vietnam

www.ssmrmh.ro

Solution 1 by Sladjan Stankovic-Skopje , Solution 2 by Marian Dincă –

Romania , Solution 3 by Myagmarsuren Yadamsuren-Darkhan-Mongolia


Solution 1 by Sladjan Stankovic-Skopje

− = −( − − ) ≤


( + + )( + − )( − + )(− + + ) =

= ( + + ) − − − =

= ( + − ) ( )

Let: + − = , + − = , + − =

Rezult: = + , = + , = +

( + − ) = ( + + )

= ( + )( + ) = ( + + + ) ≥

≥ ( + + ) ⇔ ≥


(( + ) − ) ⋅ ( − ( − ) ) =

= − + (( + ) + ( − ) ) ⋅ − ( − ) ≤ ⇔

ASSURE

⇔ − (( + ) + ( − ) ) ⋅ + + ( − ) =

= − ( + ) + ( + ) = − ( + ) ≥


− ≤ ⇔ − − − + + − ≤

www.ssmrmh.ro

⇔ + + + − ( + ) ≥

⇔ − ( + ) + ( + ) ≥ ⇔ ( − − ) ≥ ,

which is true (Proved)

40. Prove that if , , , , , ∈ ( ,∞) and + + = ; + + =

then: ⋅ ⋅ ≤




Solutions 3,4 by Uche Eliezer Okeke-Anambra-Nigeria


Probar para todo , , , , , ∈< 0,∞ >, de tal manera que

+ + = , + + = lo siguiente ≤

Por la desigualdad ponderada ≥

⋅ + ⋅ + ⋅+ + ≥ =

= ≥ (LQQD)


≤⏞

= = =


If , , , , , ∈ ( ,∞); + + = ; + + =

www.ssmrmh.ro

⋅ ⋅ ≤ ; = ⋅ ⋅

≤ = = = (Proved)


= ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅

≤+ ++ +

⋅+ +

( + + )

= ( + + ) ⋅ ⋅ ≤( + + ) ( + + )

⋅ ⋅

= ⋅ ⋅ ⋅ = (Proved)

41. Prove that:

+ + + < 4


Solution 1 by Rovshan Pirguliyev-Sumgait-Azerbaidian , Solution 2 by Uche

Eliezer Okeke-Anambra-Nigeria , Solution 3 by Soumava Chakraborty-

Kolkata-India , Solution 4 by Myagmarsuren Yadamsuren-Darkhan-Mongolia

Solution 1 by Rovshan Pirguliyev-Sumgait-Azerbaidian

= , = , = . Ineq ⇔ + < 4. It is known that

< , < we have

www.ssmrmh.ro

+ < + = = ( ) =

=( + ) +

=+

<

<+

=⋅ + ⋅

=


Let = ; = ; = . = +

= + ≤⏞ + = (*)

Now < ; 3 < ; < . So < ; < 2; < 2

⇔ (*) < ( ) ( ) ( ) = < 4. (Proved)


Let = , = , = . We have: < <

∴ = + . Now, <⏞ √ <⏞( )

√ = (∵ < )

Again, ∑ >⏞ ( ) ⇒ ∑ <( )

⇒ ∑ <( )

= √ <⏞( )

√ = (∵ < , < )

(1), (2) ⇒ < + = + = <

= = < 4 (∵ < ) = (Proved)


===

+ + = <+ = <

(*)

www.ssmrmh.ro

++

+ +=

++ ⋅

+ +≤⏞

≤ + + + + =

= ( + ) + ( + + ) <⏞(∗)

⋅ + =

42. From the book: “Math Accent”

If , , ∈ ℕ − { , }, + + = then:

+ + + + + > 200





∴ + =( + )!

! ! =( + ) … ( + )

!

=[( + ) … ( − + )]

[ . … ]( − + )( + )

Also, ≥ ⇒ (∀ = , … , ): − + ≥ ⇒ [( )…( )][ . … ] ≥ .

So + ≥ ( )( ) (*)

Then ∑ + ≥ ∑ ( )( ) > ∑ ( ) = ∑ =

A Generalization of Daniel Sitaru’s binomial inequality

Let , … , ≥ ∈ ℕ such that + ⋯+ = + for ≥ .

www.ssmrmh.ro

So, +

++

+ ⋯++

> 66 + .

For a proof, see my proof in the case = .


For ≤ ≤ − , ≥ . As , , ≥ , + ≥ +

+ ≥ + , + ≥ +

∴ + + + + + > 2( + + ) =

43. If ∀ ∈ ℝ,+ ≥ +

+ + ≥ + ++ + + ≥ + + +

, , , , > 0 then:

⋅ ⋅ ⋅ =



Solution 2 by Șerban George Florin – Romania


Let the function : → , ( ) = ∑ . Observe that ( ) = ∑ .

From the Fermat’s mean value theorem, since ( ) ≥ ( )∀ ∈ , we

get ( ) = ⇒ ∏ = .

So the first condition gives us = , the second gives = and

the third = . Multiplying, get .

Solution 2 by Șerban George Florin – Romania

( ) = + , + ≥ + (∀) ∈ ℝ

⇒ ( ) ≥ ( ) ⇒ = minimum point. ( ) = +

www.ssmrmh.ro

T. Fermat ⇒ ( ) = = +

( ) = + + , + + ≥ + + (∀) ∈ ℝ

( ) ≥ ( ) ⇒ = minimum point T. Fermat ( ) = ( ) = + + =

( ) = + + + , + + + ≥ + + + (∀) ∈ ℝ

( ) ≥ ( )|(∀) ∈ ℝ ⇒ = minimum point

T. Fermat ⇒ ( ) = = + + + = ⇒

= ; ≠ ⇒ = ⇒ = ⇒ = , ≠ ⇒ = ⇒ =

+ = ⇒ ⋅ = ⇒ ⋅ = ⇒ ⋅ =

⇒ ⋅ ⋅ ⋅ = ⋅ ⋅ =

44. In :

+ + ≥

+ + ≥





Probar en un triàngulo

1) + + ≥

2) + + ≥

Saberes previos

www.ssmrmh.ro

+ + = , + + =

Por la desigualdad de Cauchy

+ + ≥+ +

+ +=

+ + ≥+ +

+ +=


Let = , = and = .

Obviously , , > 0 and + + = .

By Cauchy, + + ( + + ) ≥ ( + + ) ⇒

+ + ≥ ⋅ ( + + ) =

The second is identical, with the specification that + + = .

45. If , ∈ ℝ , < then:

√ + − √ +−

<+ √ ++ √ +


Solution by Rozeta Atanasova-Skopje

Let = and = .

( ) = and ( ) = and > ⇒

− < − … (1)

www.ssmrmh.ro

= and ( ) = < 1,∀ ∈ ℝ ⇒

⇒ < … (2)

=√ + − √ +

− =( − )

− <( ) ( − )

+

=⋅

= <( )

− = −

= + + − + + =+ √ ++ √ +

=

46. If , > 0; ∈ ℝ then:

( + )( + )( + ) + + ≥



Prakash-New Delhi-India, Solution 3 by SK Rejuan-West Bengal-India, Solution

4 by Seyran Ibrahimov-Maasilli-Azerbaidian


Siendo , > 0, ∈ . Probar que

( + )( + )( + ) + + ≥

De las condiciones se puede afirmar lo siguiente

= + > 0; = + > 0

⇒ + = ( + ) + ( + ) = +

La desigualadad es equivalente

www.ssmrmh.ro

⇒ ( ) + + ≥ (Lo cual es cierto por ≥ )


Put = + > 0

= + > 0 ⇒ + = +

∴( + )

( + )( + ) + +

=( + )

+ + = + + + + ≥

∵ + ≥ ∀ , > 0


Given that, , > 0, ∈ ℝ

Now, ( + ) = ( + ) + ( + )

= ( + ) + ( + )

Hence by AM ≥ GM we get,

+=

( + ) + ( + )

≥ ( + )( + )

⇒ ( )( )( ) ≥ (1)


= =

++

+≤ +

+≤

+≤

www.ssmrmh.ro

+ ≥ +

≥ ⇒ + ≥ ( + ) (Proved)

47. Let be a tetrahedron with ∠ = ∠ = ∠ = ° and let

be any point inside the triangle . Denote respectively by , , the

distances from to faces ( ), ( ), ( ). Prove that

(a) + + = .

(b) ≤ ⋅ ⋅

(c) ⋅ + ⋅ + ⋅ ≥ .



Let’s take ⃗ = ̂, ⃗ = ̂, ⃗ = , where , , are positive.

Equation of plane is + + = .

Let ( , , ) be any point inside , then

+ + + = ,< < ,< < ,< <

Note that = , = , =

Now, = , = , = .

(a) + + = + + =

(b) = + + ≥ ⋅ ⋅ ⇒ ≤

(c) ( )( ) + ( ) + ( )( ) =

= + + + +

www.ssmrmh.ro

= + + + + + + + +

≥ + + + ⋅ + ⋅ + ⋅

= + + =

48. If , , , > 0 then:

( + ) ( + ) ( + ) ≤ ⋅ ⋅ ( + + + )


Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by Soumitra

Mandal-Chandar Nagore-India


Siendo , , , > 0. Probar

( + ) ( + ) ( + ) ≤ ⋅ ⋅ ( + + + )

⇔+ + +

+≥

+ +

⇔+ + +

+≥ + +

Por la desigualdad ponderada MA ≥ MG

≥ + + ⇔ ≥ + +

⇔+ + +

+≥ + +


Applying Weighted AM ≥ ;

www.ssmrmh.ro

⋅ + + ⋅ + +

≥+

⋅+

⇒+ + +

+≥

+⋅

+

⇒ ( + + + ) ⋅ ⋅ ≥ ( + ) ⋅ ( + ) ⋅ ( + )

49. If ∈ ( , ); > 0 then:

+ + + ≤√

+


Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by Seyran

Ibrahimov-Maasilli-Azerbaidian, Solution 3 by Soumitra Mandal-Chandar

Nagore-India


Si ∈< 0, >, > 0. Probar que

+ + + ≤√

+

Dado las condiciones se puede afirmar que > 0, > 0

Aplicando MA ≥ MG

+ + + ≤ +√

= +√

Además por MA ≥ MG

√+

√+ + ≥ = (A)

www.ssmrmh.ro

+ +√

≥√

=√

(B)

Sumando (A) + (B) √

+ ≥ +√

⇔

⇔√

+ ≥ +√

≥+

++


= = ; + ≤ + ; ≤

≤ ; + ≥ + ; ≥ ⇒ + ≥ ( + )


++

+≤⏞

√+

We need to prove, +√

≥√

+

⇔ +√

−√

≥ , which is true

∵ > 0, > > 0, > 0 Hence proved

50. Prove that if , ∈ ( ,∞) then:

( ) ≤+ √ ++ √ +



Let ( ) = + √ + − + , ≥

( ) =+ √ +

+√ +

− + , > 0

www.ssmrmh.ro

=√ +

+ ( + ) − > ( + ) ( + ) − = ∀ > 0

∴ ( ) increases on [ ,∞) ⇒ ( ) > ( ) ∀ > 0

⇒ ( ) > 0∀ > 0 ⇒ 6 + + > 6 − ∀ > 0

⇒ + √ + > ∀ > 0 (1)

Next, let ( ) = − √ + + + , ≥

( ) = −√ +

+ + , > 0

⇒ ( ) = −√ +

+ > 0∀ > 0

∴ ( ) increases on [ ,∞) ⇒ ( ) > ( )∀ > 0

⇒ − + > −6 − ∀ > 0

⇒ −√ + > ∀ > 0 (2)

Putting = in (1), = in (2), (with , > 0) we get

− + + + > ⋅

⇒+ √ +

+ √ +> ( ) ,∀ , > 0

51. If , , , ∈ ℝ then:

+ + + ( + + )( + + ) ≥


Solution 1 by Ravi Prakash-New Delhi-India, Solution 2 by Boris Colakovic-

Belgrade-Serbia, Solution 3 by Bedri Hajrizi-Mitrovica-Kosovo

www.ssmrmh.ro


= ( + + ) + ( + + )( + + ) ≥

If + + ≥ , then write

= ( + + − − − ) + ( + + )( + )( + )

= ( − ) + ( − ) + ( − ) + ( + + )( + )( + ) ≥

If + + < 0 and + + < 0, still ≥ .

If + + < 0 and + + > 0,

then + + ≤ √ + < 2 Now, write

= ( + + ) + ( + + )

{( + + ) − − − }

= ( + + )( − − − ) +

+( + + ) ( + + ) ≥

⇒ ≥ in this case. Thus, ≥ ∀ , , , ∈ ℝ

Solution 2 by Boris Colakovic-Belgrade-Serbia

+ = , =−

⇒ + + =

= + − =( + )

− ; + + ≤ + + ⇔

( + )− ( + + ) ≤

( + )− ( + + ) ⇔

⇔ ≤ + + + ( ) − ( + + ) ≤ ( ) ( + + )

Solution 3 by Bedri Hajrizi-Mitrovica-Kosovo

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

( + + ) + ( + + )∈ ,√

( + + ) ≥

www.ssmrmh.ro

It’s enough to prove that: ( + + ) − ( + + ) ≥

+ + ≥ + + which is true for all , , ∈ ℝ

52. :ℝ → ℝ, ( ) = , , sides in scalene

Prove that:

( )( ) >



( ) =

As ( ) = when = , = , = , = , = , =

( − ), ( − ), ( − ), ( − ), ( − ), ( − ) are factors of ( )

Also, ( ) is a homogenous expression of degree in , , ,

∴ ( ) ≡ ( − )( − )( − )( − )( − )( − )( + + + )

where is a constant. When = ,

( ) = − = −− −− −− −

= −( − )( − ) + ++ + + + + +

→ − , → −

www.ssmrmh.ro

( ) = −( − )( − )+ +

= − ( − )( − )+ +

= − ( − )( − )+ +

= − ( − )( − )( − )( + + )

Also, ( ) = ( − )( − )( − )(− ) ( + + )

= ( − )( − )( − )( + + ) ∴ = − . Thus,

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

| ( )| = |( − )( − )( − )| + | − | + | − | +

+ | − | + | + + + |

⇒( )( ) = − + − + − + + + +

( )( ) = − + − + − + + + +

≥ ( − )( − ) + ( − ) + = + = >

53. If , , ∈ ℝ∗; , , ∈ ℝ

= + +

= + +

then: + ≥



www.ssmrmh.ro

Now, + = + + + + +

( + + )

+ + + + ( + + ) + + +

= + + + + + +

+ + + + + + +

+ + + + + +

= + + + + + + + + ≥

54. If , ∈ , then:

++

++

+<

√


Solution by Khanh Hung Vu-Ho Chi Minh-Vietnam

If , ∈ ; then + +⋅

≤ √

Put = , =⋅

, =⋅

( , , > 0). We have:

+ ( ⋅ ) + ( ⋅ ) = + = ⇒

⇒ + + = . We have + ⋅ + ⋅⋅

≤ √

www.ssmrmh.ro

⇒+

++

++

≤√

⇒+

++

++

≤√

We have ≤ √ ⋅ + √ ⇒ ≤ √ √ ⋅ ⇒

⇒ ≤ √ + √ ( + )

⇒ √ ⋅ − + √ ⋅ + √ ⋅ + √ ≥ ⇒

⇒ − √ √ − − √ − − √ − ≥ ⇒

⇒ −√ √ ⋅ + ⋅ + √ ⋅ + + √ ≥

Similarly, we have ≤ √ ⋅ + √ and ≤ √ ⋅ + √

⇒ + + + + + ≤√

+ + +√

=√

=√

The equality occurs when

= = = √ ⇒ = ⋅ = ⋅ =√

We have = √ ⇒ = − √ = √ and = √

We have ⋅ = ⋅ = √ ⇒ = = √ ⇒ =

55. Prove that if , , ∈ ( ,∞); = then:

+ + ≥


Solution by Boris Colakovic-Belgrade-Serbia

+ + ≥ =

www.ssmrmh.ro

+ + ≥ + = + ≥ +

+ + ≥ = ⋅ = >

56. If , , ≥ then:

− ≥ ( − )( − )( − )


Solution by Soumitra Mandal-Chandar Nagore-India

( − )( − )( − ) + + + − − − = −

− ≥ ( − )

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

( − )( − )( − )

⇔ ( − ) {( + )( + + ) − } + ( − ) +

+ ( − ) + ( − ) ≥ , which is true since , , ≥ and

( + )( + + ) ≥ ∴ − ≥ ( − )

(Proved)

57. If , , ≥ , + + = then:

( ) + ( ) + ( ) ≥( + + )


Solution by Sanong Hauyrerai-Nakon Pathom-Thailand

www.ssmrmh.ro

+ + ≥( + + )

, , ≥ and + + =

Definition = , ∈ ℝ and − < <

Iff = , ∈ ℝ and – < <

Proof give = Iff =

= Iff = ; = Iff =

consider ≤ + + = + + ≥ ( )

Iff ( + + ) ≥ ( ) and since + + =

Hence, , , ≥ , , , ≥

and ≥ , ≥ , ≥

Hence, + + ≥ ( + + ) ≥ ( )

Therefore + + ≥ ( )

58. If , > 0, ∈ ℕ, ≥ then:

+√+

+ −√+

< 2


Solution 1 by Ravi Prakash-New Delhi-India, Solution 2 by Geanina Tudose –

Romania, Solution 3 by Sanong Hauerai-Nakon Pathom-Thailand, Solution 4

by SK Rejuan-West Bengal-India, Solution 5 by Nikola Djurici-Serbia, Solution

6 by Ngo Minh Ngoc Bao-Vietnam

www.ssmrmh.ro


We assume ≥ ; + √ = √ √ = √ √√

and − √ = √ √ = √ √√

. Note that

√ √ + √ √ = ( ) = . Put √ √√

= √ ,

and √ √√

= √ , ≤ < . Now,

= +√+

+ −√+

=√ + √√ +

+√ − √√ +

= √ + √ = ( ) where ( ) = ( ) + ( )

≤ ≤ ; ( ) = ( ) (− ) + ( ) ( )

= ( ) ( ) − ( ) ( ) ; ( ) = ⇒ =

( ) = ( ), , = ,( )

, =( )

∴ ≤

Solution 2 by Geanina Tudose –Romania

By GM ≤ AM we have ⋅ ⋅ … ⋅ ⋅ + √ ≤... ⋯ √

⋅ … ⋅ ⋅ −√+ ≤

+. . . + + + √+

www.ssmrmh.ro

Summing up, we have + √ + − √ ≤√ √

=

The inequality is strict, since ± √ ≠ , , > 0

Solution 3 by Sanong Hauyrerai-Nakon Pathom-Thailand

, > 0, is integer ≥

+ ≤ ( + )

+ ≤ ( + )

+ ≤ ( + )

…

…

+ ≤ ( )( + )

hence + ( )( ) + −

≤ ( ) + + + = ( − )( ) = ( ) =


If , > 0, ∈ ℕ, ≥ , then, by mth power theorem we get,

+ √+ + − √

+<

+ √+ + − √

+

www.ssmrmh.ro

⇒ +√+

+ −√ +

< 2 =

⇒ +√+ + −

√+ < 2

Solution 5 by Nikola Djurici-Serbia

≤ weighted mean inequality. Put = + ( )( ) ,

= −( )+ ,

+= ; ⋅ = ⋅ =

Equality holds only if = , but that would mean that = or = , so

equlity isn’t reached, so it’s strict <.

Solution 6 by Ngo Minh Ngoc Bao-Vietnam

Let = √ ≤ √√

= . Considering function

( ) = ( + ) + ( − ) ,∀ ∈ ( , ]

⇒ ( ) = ( + ) − ( − ) < 0,∀ ∈ ( ; ] ⇒ ( ) < ( ) =

59. If , , ≥ , + + = then:

√ + √ + √ ≤



Prakash-New Delhi-India, Solution 3 by Togrul Ehmedov-Baku-Azerbaidian


Siendo , , ≥ , de tal manera que + + = . Probar que

www.ssmrmh.ro

√ + √ + √ ≤ . Para ello probaremos que

+ + = ⇔ + + =

Partimos de + = ⇔ = − ⇔ ( ) = ( − )

⇔ = − ⇔−

− = −−

− +

⇔ ( − )( − + ) =

= ( − )(− + )

⇔ − + − + − =

= − + −

⇔ − + − = , cuyas raíces son = ,

donde → = , , , , , ,

⇔ ( − + − ) =

⇒ − + − = , cuyas raíces son = ,

donde → = , , , , ,

Ahora bien llevando a “ ” , donde ≠

⇔ ( − + − ) =

⇔ − + − =

Haciendo =

⇔ − + − = , cuyas raíces son

= , ,

Aplicando Vietta ⇒ + + = − =

Aplicando Cauchy en la desigualdad propuesta

www.ssmrmh.ro

√ + √ + √ ≤

≤ ( + + ) + + ≤

≤ ( + + ) = √ =


Let = , = ⇒ = ( − )

⇒ − =

, , are roots of ( − ) = ( − )( − )

⇒ − + − =

+ + =∑

= =

Now, by C-S inequality √ + √ + √ ≤

≤ √ + + + +

≤ + + √ = √ √ =

Solution 3 by Togrul Ehmedov-Baku-Azerbaidian

√ + √ + √ ≤ ( + + ) + +

≤ ( + + ) + + ≤ + +

= + + = + +

www.ssmrmh.ro

=+ +

=− + − + −

=∑ − + +

=− + +

=

−

=+

=√

=

√ + √ + √ ≤

√ + √ + √ ≤

60. If , ∈ ℝ, ≠ then:

( + + ) − ( + + )−

> ⋅ √


Solution 1 by Chris Kyriazis-Greece,


Solution 1 by Chris Kyriazis- Greece

The function ( ) = + +

www.ssmrmh.ro

∈ ℝ is strictly convex and positive for every ∈ ℝ (easy)

So, using the Hermite – Hadamard inequality we have: (if > )

∫ ( )

−>

+⇔

+ + − ( + + )−

> + + =

= + ⋅ +

> ⋅ ⋅

= ⋅ = ⋅ √


WLOG > . We have: ( ) ( ) = ( ) ( ) ( ) ( ) =

=( − )( + + + + + )

−=

=( − ) ⋅ ( + + + + + )

( − ) ⋅

⇒ ( ) ( ) = ( )⋅ ( )

( )⋅ (1)

We have

+ + + + + ≥ + √ ⋅ + √ ⋅ + =

= + ⋅ + ⋅ + = + ⋅ + ⋅

www.ssmrmh.ro

On the other hand by AM-GM we have

+ ⋅ + ⋅ = + + + + + ≥

≥ ⋅ ⋅ ⋅ ⋅ ⋅ = √ = ⋅

⇒ + + + + + ≥ ⋅ ⇒ ( + + + + + ) ≥

≥ ⋅ ⋅ = ⋅ ⋅( )

= ⋅( )

= ⋅ √ (2)

(1), (2) ⇒ ( ) ( ) ≥ ( )⋅ ⋅ √

( )⋅ (3)

We need to prove that ( )⋅ ⋅ √

( )⋅> √ (4)

We have (4) ⇒ ( )⋅ √

( )⋅> 6 ⋅ √ ⇒

( )⋅> (5)

Put = ( > 0). We have (5) ⇒⋅

> ⇒ − > ⋅ ⋅ ⇒

⇒ − ⋅ ⋅ − > 0 (6)

( ) = − ⋅ ⋅ − ⇒ ( ) = ⋅ − ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅

⇒ ( ) = ⋅ − ⋅ − ⋅ ⋅ = ⋅ ( − ⋅ − )

( ) = − ⋅ − ⇒ ( ) = ⋅ − > 0 ⇒ ( ) is a increasing function

⇒ ( ) > ( ) = ⇒ ( ) > 0 ⇒ ( ) is a increasing function ⇒

⇒ ( ) > ( ) = ⇒ (6) true ⇒(5) true ⇒ (4) true

(3), (4)⇒ ( ) ( ) > ⋅ √

Q.E.D.

61. If , ∈ , then:

( + ) < +


Solution by Soumava Chakraborty-Kolkata-India

www.ssmrmh.ro

Let’s prove that ∀ ∈ , , >

( ) > ⇔ >

Let ( ) = − ( ) =

( ) = + ( )−

= + − = ( ) ( ) =

( ) = + − = ( ) ; ( ) =

( ) = ( ) + ( )( )( ) + ( − ) −

= ( + ) + ( + ) + ( − ) −

= ( + ) + ( + ) ++

− =( + ) + ( + ) + − ( + )

+

=+ + ( ) + + ( ) + + − ( )

+=

++

> 0,

( = > 0)

∴ ( ) > 0 and ( ) = ⇒ ( ) > ℎ( ) = ⇒ ( ) > 0 and ( ) =

⇒ ( ) > ( ) ⇒ ( ) > 0 and ( ) = ⇒ ( ) > ( ) =

⇒ > ⇒ > ∀ ∈ ,

∴ > and >

⇒ + > + = ( + )

62. If ∈ ℕ, ≥ then:

+ + + + > 2 +


www.ssmrmh.ro


If ∈ , ≥ then

+ + + + > 2 +

We have

> = ⇒ + + + > 1

⇒ + + + + > + 1 ≥ 3

⇒ + + + + > 3 (1)

On the other hand, we have ≥ (Since ≥ )

⇒ ≤ ⇒ + ≤ (2)

(1) and (2) ⇒ + + + + > 2 +

63. If , , … , > 0, ∈ ℕ∗, :ℝ → ℝ, ( ) = + +⋯+ then:

+ + ⋯+ + ≥( + )


Solution 1 by Nirapada Pal-Jhargram-India,Solution 2 by Soumitra Mandal-

Chandar Nagore-India, Solution 3 by Ngoc Minh Ngoc Bao-Gia Lang-Vietnam


= ⋅∑

≥

⎝

⎛∑

⎠

⎞

www.ssmrmh.ro

=

⎝

⎜⎛∑ ∑

⎠

⎟⎞

As ( ) = ∑

=

⎝

⎜⎜⎛

⎝

⎜⎛∑

⎠

⎟⎞

⎠

⎟⎟⎞

≥⏞

= =( + )


≥

= + + + ⋯+

≥⏞ ( + + + ⋯+ ) = ( ) (proved)

Solution 3 by Ngoc Minh Ngoc Bao-Gia Lang-Vietnam

Use Cauchy – Schwarz inequality:

+ + ⋯+ + ≥ + + ⋯+ + (*)

We have:

+ + ⋯+ + = +⋯+ +

www.ssmrmh.ro

= + + ⋯+ + + + + ⋯+ + + ⋯

… + + + ⋯+ +

≥ ⋅ ⋅… ⋅ ⋅ + ⋅ ⋅… ⋅ ⋅ +⋯

+ ⋅ ⋅… ⋅ ⋅ =( + )

⇒ (*) ≥ RHS (*) ≥ ⋅ ( ) = ( )

Equality when = = ⋯ =

64. If = + , = + , = + ,

, , = , then: √ + √ + √ ≤ √

Proposed by Boris Colakovic-Belgrade-Serbia

Solution 1 by Nirapada Pal-Jhargram-India, Solution 2 by Abdul Aziz-

Semarang-Indonesia, Solution 3 by Myagmarsuren-Yadamsuren-Darkhan-

Mongolia


+ + = gives ∑ =

Now, ∑√ = ∑√ ≤ ∑ = √ ∑ + = √


Since + + = then + + =

⇔ + + − = ⇔ + + =

By CS, √ + √ + √ ≤ ( + + )( + + )

www.ssmrmh.ro

⇔ √ + √ + √ ≤ √ ⋅ = √

Solution 3 by Myagmarsuren-Yadamsuren-Darkhan-Mongolia

If

= ⋅ += ⋅ += ⋅ +

; ; > 0+ + = ⎭

⎪⎬

⎪⎫

then √ + √ + √ ≤ √

( ) = √ ⇒ ( ) ≤

⇒ √ + √ + √ ≤ ⋅ ∑ ⋅ =

= ⋅+ ∑ ⋅

=⋅ √

= √

65. If in , ≥ ≥ then:

+ + <



≥ ≥ ⇒ ≤ ≤ ⇒by Chebyshev’s sum inequality

≤ ( + + ) + + (1)

But + + < + + (2)

because ( ) = + − ( + )

= + + < ( + ) ⇒ < + , and similarly

+ < +

< +

www.ssmrmh.ro

− − − − −− −

( + + ) < 2( + + )

On the other hand

+ + = + + + (3)

Let’s consider the solutions of − = ⇒

= ⇒ ( ) = ⇒

= + + + + + +

= + + + + + = − (4)

From (1), (2), (3) and (4) ⇒

< ( + + ) − = ⋅ = =

66. If , , ∈ ( ,∞), = then:

+ ( + )+ >



If , , ∈ ( , +∞) and = . Prove that

⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ >

Lemma: > if ∈ ;

Since ∈ , and ∈ , , applying the lemma, we have:

> ⋅ = and > ⋅ =

www.ssmrmh.ro

We have ⋅ + ( + ) ⋅ > ⋅ + ( + ) ⋅ =

= + ( + ) = + > 2√ ⋅

Similarly, we have ⋅ + ( + ) ⋅ > 2√ ⋅ and

⋅ + ( + ) ⋅ + > 2√ ⋅

So

⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ >

> 2√ ⋅ ⋅ √ ⋅ ⋅ √ ⋅ = √ ⋅ ⋅

⇒ ⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ ⋅ + ( + ) ⋅ >

> 8 ⋅ ⋅ =

67. If ∈ ℕ∗, ≥ , , , > 1, + + = then:

+ √ + − √ < 18


Solution 1 by Khanh Hung Vu-Ho Chi Minh-Vietnam, Solution 2 by SK Rejuan-

West Bengal-India, Solution 3 by Eliezer Okeke-Anambra-Nigeria


If ∈ ℕ∗, ≥ , , , > 1, + + = then

+ √ + − √ + √ + − √ + √ + − √ < 18

By AM-GM, we have:

+ √ = √ √ +√

≤( )⋅ √ √

√ =

www.ssmrmh.ro

= √ +√

Similarly, we have − √ ≤ √ −√

⇒ + √ + − √ ≤ ⋅ √ (1)

On the other hand, by AM-GM we have ⋅ ( ) ≤ ( )⋅

⇒ √ ≤ ( )⋅⋅ ( )

= ( )⋅⋅

(2)

(1), (2) ⇒ + √ + − √ ≤ ( )⋅⋅

Similarly, we have + √ + − √ ≤ ( )⋅⋅

and

+ √ + − √ ≤+ ( − ) ⋅

⋅

⇒ + √ + − √ + + √ + − √ + + √ + − √ ≤

≤+ ( − ) ⋅

⋅ ++ ( − ) ⋅

⋅ ++ ( − ) ⋅

⋅

⇒ + √ + − √ + + + − + + √ + − √

≤

≤( + + ) + ( − ) ⋅

⋅ =⋅ + ( − ) ⋅

⋅ =( + ) ⋅

⋅ =( + )

Since < 2 ⇒ ( ) < 18

So + √ + − √ + + √ + − √ + + √ + − √ ≤

The equality doesn’t exist. Therefore,

+ √ + − √ + + √ + − √ + + √ + − √ < 18

www.ssmrmh.ro


If ∈ ℕ∗, ≥ , , , > 1, + + =

By mth power theorem we get = as, √ < ⇒ − √ > 0

+ √ + − √<

+ √ + − √

⇒

⎩⎪⎨

⎪⎧

+ √∥

( )

+ − √

⎭⎪⎬

⎪⎫

< 2 =

⇒ < 2∑ (1)

Again by mth power theorem, ∑ < ∑ = =

[∵ ∑ = ] ⇒ ∑ < 3 ⇒ ∑ < 9 ⇒ 2∑ < 18 (2)

Combining (1) & (2) we get,

< 2 < 18 ⇒ + √ + − √ < 18

Solution 3 by Eliezer Okeke-Anambra-Nigeria

for , , > 1 + + = ≥ ∈ ℕ

prove ∑ + √ + − √ < 18. Consider the composite

functions ( ) = + √ and ( ) = − √ both functions are

concave. Hence ( ) = ( ) + ( ) is concave.

Lemma: √ < √ for > 1 (1)

www.ssmrmh.ro

Lemma: AM ≤ AM (2)

We apply Jensen

+ √ + − √ ≤⏞

⎣⎢⎢⎢⎡ ∑

+∑

+∑

−∑

⎦⎥⎥⎥⎤

=

⎣⎢⎢⎢⎡

+ + −

⎦⎥⎥⎥⎤

= ( + ) + ( − ) <⏞( )

( + ) + ( − )

<⏞( ) ( + ) + ( − )

= ( ) =

68. If , , , > 0, + + + = 1 then:

+ + + + ( + + + + + ) ≥

≥ + √ + √



Siendo , , , > 0, de tal manera que + + + = . Probar que

+ + + + ( + + + + + ) ≥

≥ + √ + √

Aplicando la siguiente identidad conocida

( + ) = + + ( + ), donde = + , = + ⇒ ( + + + ) = ( + ) + ( + ) + ( + + + )( + )( + )

⇒ = + + ( + ) + + + ( + ) + ( + )( + )

⇒ + ( + ) + ( + ) = + + + + ( + + + ) +

www.ssmrmh.ro

+ ( + + + ) + ( + )( + )

⇒ + ( + ) + ( + ) = + + + + + + ( + + + )

⇒ + ( + ) + ( + ) = + + + + ( + + + + + )

Como , , , > 0. Aplicando ≥

⇒ + + + + ( + + + + + ) =

= + ( + ) + ( + ) ≥ + √ + √

69. If ∈ ℕ∗, > 0 then:

!

(− ) −+ < +



Consider the expression = ∏

We split this expression into partial fractions.

= (− + )( + ) … (− )( ) … ( + − ) ⋅ +

=(− )

!⋅

!( − )! ( + − )!

⋅+

=!∑ (− ) − . Also, + + > 1

∴!

(− ) − += <

+≤

+

⇒!

(− ) − +<

+

www.ssmrmh.ro

70. If , ≥ , ∈ ℕ∗ then:

⋅ ⋅ ≥ ( − )



∑ ⋅ ⋅ ≥ ( − ) (1)

If = or = , there is nothing to prove. ∴ suppose , > 0

Now, (1) can be written as

≥ − ⇔ ≥ −

If is odd,

≥ + −

= + ≥ = = −

If is even, ∑

≥ + − +

= + +

www.ssmrmh.ro

≥ + = = −

71. If , , > 0, + + + = 0 then:

| + + + | ≥ +


Solution 1 by Kevin Soto Palacios-Huarmey-Peru, Solution 2 by Ravi Prakash-

New Delhi-India

Solution 1 by Kevin Soto Palacios-Huarmey-Peru

Siendo , , > 0, de tal manera que + + + = . Probar que

| + + + | ≥ +

De la condición ⇒ ( + ) = −( + ) ⇔ ( + ) + ( + ) =

⇒ + + + + ( + ) + ( + ) =

⇒ + + + − ( + ) − ( + ) =

⇒ + + + = ( + ) + ( + )

Por lo tanto, Como , , > 0; Aplicando MA ≥ MG

| + + + | = + + + ≥ +


Let , , > 0, + + + = 0

− = ( + + ) ≥ ( ) ⇒ − ≥

⇒ − − ≥ ⇒ + ≤ − < 0

Also, + + + = (− − − ) + + +

= −( + ) + ( + ) < 0

www.ssmrmh.ro

∴ the given inequality becomes

− ( + + + ) ≥ − − . Now,

− = (− ) ≥ + + (− ) ⇒ − ≥ + −

Similarly − ≥ + − . And − ≥ + − . Thus,

− − − − ≥ ( + + − )− − (1)

But ≥ ⇒ + + ≥ −

⇒ ( + + − ) ≥ − − > 0 (2)

From (1), (2) we get − − − − ≥ − −

or | + + + | ≥ +

Equality when = =

72. If , , ≥ − then:

− ≥ ( − )


Solution by Richdad Phuc-Hanoi-Vietnam

WLOG, assume ≤ ≤ or ≥ ≥ . We have

− = ( − ) ( + ) − ( + ) + ( − ) − − +

Let ( ) = ( + ) , ≥ − , ( ) = ( + ) > 0,∀ > −2

is increasing function on [− , +∞)

⇒ ( − ) ( + ) − ( + ) ≥ ,∀ , ≥ −

Let ( ) = , ≥ − , ( ) = ( + ) > 0, for all ≥ −

is increasing function on [− ; +∞) case ≤ ≤

www.ssmrmh.ro

≥≥

⇒ ( − ) − + − ≥

we get − ≥ case ≥ ≥ ≤≤

− ≥ ⇒ . . . Equality hold if = =

73. If < < then:

>



∀ > 1: + ≤ ⇔ ( − ) − ≤ ⇔( )

( ) ≤ ⇔ ≤

< < ⇒ ∫ > ∫ ⇔ > ⇔ >

74. Prove that + ≥√√ ,∀ ∈ ℝ.

Proposed by Ibrahim Abdulazeez-Zaria-Nigeria

Solution by Daniel Sitaru-Romania

+ ≥⏞ = =

= √ ≥⋅√ ≥ √ =

√√

75. Prove that:

° + ° + ° + ⋯+ ° < 54 ⋅ °

Proposed by Ilkin Guliyev-Azerbaidian

www.ssmrmh.ro


Note: ( ) ≤ ∀ ∈ ℕ, < < [ = , ( ) ≤ ]

Assume ( ) ≤ for some ∈ ℕ.

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

≤ ( ) + ≤ + = ( + ) ∴ ( °) ≤ °

⇒ ( °) ≤ ° = ( °)

76. If , , ∈ , then:

( + + ) ≤ + +

Proposed by Daniel Sitaru-Romania

Solution 1 by Anas Adlany-El Zemamra-Morocco, Solution 2 by Abdelhak

Maoukouf-Casablanca-Morocco

Solution 1 by Anas Adlany-El Zemamra-Morocco

∑ ( ) ≥⏞ (∑ )(∑ ( )) ≥⏞ ∏ (∑ ( )) =

= ( ) .


→ is a descending function on ; , So by Chebyshev:

≥ ⇔ ( + ) ≥

⇔ + ≥ +

www.ssmrmh.ro

if ∑ ≤ ∑ similarly we’ll have ∑ ≤ ∑

⇒ ∑ + ∑ ≤ ∑ + ∑ ∵False supposition

So ∑ ≥ ∑ ⇔ ∑ ≥ ∑

77. If ≤ , , < 1 then:

( + )( + )( + )( − )( − )( − ) ≥

+−



For ≤ , , < 1, consider

= ( − ) ( + )( + )( + ) −

−( + ) ( − )( − )( − )

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

Use → − to obtain

= ( − ) +( − )( − )( − ) ( + + ) +

Use → +

= + ++ + + + + +

Use → −

= + ++ + − + + −

Note that + − ( + ) = ( − ) > 0

⇒ + > 3 + (1)

www.ssmrmh.ro

Also, + + ≥ + + and

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

[∵ ≤ , , < 1] ⇒ ( + + )[( − ) + ( − ) + ( − ) ]

≥ ( + + )[ ( − ) + ( − ) + ( − ) ]

⇒ + + − ≥ + + − (2)

From (1), (2) we get ( + )( + + − ) ≥

≥ [ + + − ] ⇒ ≥

⇒ ( − ) ( + )( + )( + )

≥ ( + ) ( − )( − )( − )

Put = , = , = to obtain ( )( )( ) ≥ ( )

78. If , , > 0 then:

√+ √

++ + √

( + + ) ≥


Solution by Nguyen Thanh Nho-Tra Vinh-Vietnam

∑ √√ √

= ∑√ √

≥ √ √ √

√ √ √=

√ √ √+

⇒ =√ √ √

+ √ √ √( ) + ⇒ ≥⏞ + =

79. If , , , > 0 then:

≥ √ ≥


www.ssmrmh.ro


Consider ∑ − = ∑[( − ) + ( − ) + ( −

−ܖ܉ܜ

= ( − )( − )

As is increasing on ( ,∞), ≥ ⇒ ≥

∴ ( − )( − ) ≥ ,∀ > 0, > 0

Thus, ∑ ≥ ∑ ≥ ∑( ) . Next,

( ) ≥ ( )

= [ ]

80. If , , ∈ ℝ, > 0, | | ≤ , | | ≤ , | | ≤ then:

( − ) + ( − ) + ( − ) + ≤


Solution by Le Minh Cuong-Ho Chi Minh-Vietnam

Apply AM-GM we get:

( − ) = ( − )( + ) ≤ ( ) = (1)

( − ) ≤ (2); ( − ) ≤ (3)

and: ≤ + + (4)

From (1), (2), (3), (4), we get:

( − ) + ( − ) + ( − ) + ≤

www.ssmrmh.ro

“=” ⇔ = = =

81. If , , , , , > 0, + + ≥ 3 then:

( + + )( + + )( + + ) ≥+ +


Solution by Pham Quoc Sang-Ho Chi Minh-Vietnam

We have ( + + )( + + )( + + ) ≥⏞

≥ + + = ( + + ) ≥

On the other hand, we have ≥

so ( + + )( + + )( + + ) ≥

“=” = = = and = =

82. If , , , , , ≥ , + + = then

+ + + + + ≥ + + +


Solution by SK Rejuan-West Bengal-India

Let us consider , , with the associated weights , , respectively by

AM≥GM we get, ≥

⇒ + + ≥ [∵ + + = ]

www.ssmrmh.ro

⇒ ≥ ( ) (1)

Similarly we get, ≥ ( ) (2)

and ≥ ( ) (3)

Adding (1), (2) & (3) we get,

+ + + + + ≥

≥ ( ) + ( ) + ( ) (4)

Applying AM≥HM on right side of (4) we get

++

++

+≥

+ ( + + )( + + )

= (5)

Combining (4) & (5) we get,

+ + + + + ≥ + + +

83. If , , , > 0 then:

( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ≥

≥ + √ + √ + √ + √

Proposed by Mihály Bencze-Romania


( + )( + )( + ) − + ( )

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

www.ssmrmh.ro

− + ( ) + ( ) +

= + + − ( ) + + + − ( ) ≥

[∵ ≥ ] ⇒ ( + )( + )( + ) ≥ + ( ) (1)

Similarly, ( + )( + )( + ) ≥ + ( ) (2)

( + )( + )( + ) ≥ ( ) (3)

and ( + )( + )( + ) ≥ + ( ) (4)

Multiplying (1), (2), (3), (4) we get the required inequality.


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

≥ + ⋅ ⋅ = + √

⇒ ( + )( + )( + ) ≥ + √ (1)

Similarly, ( + )( + )( + ) ≥ + √ (2)

( + )( + )( + ) ≥ + √ (3)

( + )( + )( + ) ≥ + √ (4)

(1).(2).(3).(4) ⇒ ≥

84. In convexe quadrilater:

www.ssmrmh.ro

+ + + ≥


Solution by Geanina Tudose – Romania

For convenience denote

= ; = ; = ; = ; = ; = ; = ; =

In by Sine Theorem = ⇒ = ⋅

In , = ⋅ . Thus, = ⋅⋅

= ⋅⋅

Similarly, = ⋅⋅

; = ⋅⋅

; = ⋅ . The inequality becomes

⋅ + ⋅ + ⋅ + ⋅ ≥ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =

85. In cyclic quadrilateral, , , , – sides, - circumradius:

√( + + + ) ≤

√

Proposed by Adil Abdullayev-Baku-Azerbaidian

www.ssmrmh.ro

Solution 1 by Kevin Soto Palacios – Huarmey – Peru, Solution 2 by

Myagmarsuren Yadamsuren-Darkhan-Mongolia, Solution 3 by Soumitra

Mandal-Chandar Nagore-India


Probar en un cuadrilátero cíclico con lados , , , y –

Circunradio

( ) ≤√

. Aplicando ≥ ⇔ ( ) ≤ ≤√

Es suficiente probar

⇔ √ √ ≥ ⇔ ≥ ≥ ≥

www.ssmrmh.ro

Nuevamente por ≥ ≤ ( ) = ⋅ ≤

Usando Ptolemy theorem

⋅ + ⋅ = ⋅ ⇔ + = ⋅

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

=( )

=( ) + ( )

,

=( )

=( ) + ( )

⋅ =( ) + ( ) ( ) + ( )

≤


√≥ ( ) (1)

⇒ = ; ⇒ =

+= + ≤ ⋅

+

+ ≤ ⋅

+ ≤ ⋅

+ ≤ ⋅

+ ≤ ⋅

⎭⎪⎪⎬

⎪⎪⎫

( + + + ) ≤ + + + =

≤ + + + ≤

≤+

++

= ⋅ ⋅ = √

≤ √ ⇒√≤ (*)

www.ssmrmh.ro

(1)⇒√≥

√⇒

√≥(∗)

≥√

⇒

⇒ ≥ √ ⇔ ≥ √ ⇒+ + +

≥ √


A cyclic quadrilateral with sides , , , and semi-perimeter and

circum – radius is given by

=( + )( + )( + )( − )( − )( − )( − )

∴ ≥⏞ √ ⋅ √ ⋅ √

− + − + − + −

= ⋅√

( + + + ) ⇒√

( + + + ) ≤√

86. In tetrahedron, , , , − altitudes, − circumradius,

− inradius: ( + + + ) ≥



A Simple Proof of Euler’s Inequality in Space

Zhang Yun – Jinchang City - Gasu Province – China

Let be the radius of the circumscribed sphere of a tetrahedron and let

be the radius of the inscribed sphere of the tetrahedron. Then Euler’s

famous inequality in space state that

www.ssmrmh.ro

≥ (1)

We give here a simple proof of this inequality.

Let be the circumcenter of the tetrahedron .

Let ( = , , , ) denote the area of the face opposite the vertex ,

let denote the distance from to its opposite face, and let denote

the distance from the point to the face opposite . Then

+ ≥ , and so + ≥ . Thus, + ≥ .

Adding the four inequalities, we obtain that

( + + + ) + + + + ≥

≥ + + + .

Let denote the volume of the tetrahedron .

Then = = ( + + + ), so

( + + + ) + ≥ × , from which it follows that

( + + + ) ≥ . Since = ( + + + ),

this gives ( + + + ) ≥ × ( + + + ).

Thus, ≥ , so the inequality (1) is proved.

In two dimensions rather that three, if is now the radius of the

circumscribed circle of a triangle and the radius of the inscribed circle,

then, by a similar argument, ≥ .


Si es un tetraedro, donde , , , , son las alturas, y r el

circunradio e inradio.

Probar que ( + + + ) ≥ .

Es un tetraedro se cumple la siugiente identidad y desigualdad

www.ssmrmh.ro

+ + + = , ≥

Ahora bien, por la desigualdad de Cauchy

+ + + = + + + ≥( + + + )

+ + +=

Por lo tanto ( + + + ) ≥ ⋅ =

87. If in – tetrahedron, , , , - altitudes, − inradii then:

−+ +

−+ +

−+ +

−+ ≥

Proposed by D.M. Bătinețu – Giurgiu and Neculai Stanciu – Romania


Si es un tetraedro, donde , , , , son las alturas y el

inradio. Probar que + + + ≥

Recordar la siguiente identidad + + + =

La desigualdad propuesta es equivalente −+

+ +−+

+ +−+

+ +−+

+ ≥ +

+ + + ≥ . Aplicando la desigualdad de Cauchy

++

++

++

+≥

( + + + )

+ + + +=

88. If in - tetrahedron = = , = = , = = ,

www.ssmrmh.ro

- radii of circumsphere then:

( − )( − )( − ) ≤



Equifacial tetrahedrons exist only when the faces are congruent acute

triangles, and then = ⇒ > 0, > 0, > 0 ⇒

= ( − )( − )( − )

= ⋅+ −

⋅+ −

⋅+ −

= ( + − )( + − )( + − ) =

= ≤ + +

≤ + +

= = =

89. Let , , > 0 such that: + + = . Find the maximum of

expression: = + +

Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam

Solution by Hoang Le Nhat Tung – Hanoi – Vietnam

We have: − − + + = ( − ) + ( − ) + ( − ) − ( − ) − ( − ) − ( − )

= ( − ) + + − − −

= ( − ) ( + + + + ) ≥ ( > 0 and ( − ) ≥ )

⇒ − − + + ≥ ⇔ − + + ≥ ⇔

www.ssmrmh.ro

⇔ − + + + ≥ +

⇔ ≤ ⇔ ≤ (1)

By inequality AM – GM for 4 positive real numbers:

+ = + + + ≥ ⋅ ⋅ ⋅ = ⋅ =

= √ ⇔+

≤√

=√

Therefore (1) and by AM-GM:

⇒− + + +

≤√

≤ ⋅ + + + +

Similar: ≤ + ; ≤ +

Therefore: ⇒ = + + ≤

≤ + + + + +

⇔ ≤ + + + + + (2)

I have + + = and inequality:

( + + ) ≥ ( + + ) with: = , = , = :

= + + ≥ ⋅ + ⋅ + ⋅ = + + ⇔

⇔ + + (3)

Other let (3) and inequality AM-GM. I have:

≥ + + = + + + + + − ≥ + + − ⇔

⇔ + + ≤ (4)

Let (2), (3), (4): ⇒ ≤ ⋅ + ⋅ = = ⇒ ≤ ⇒ =

www.ssmrmh.ro

Equality occurs if:

⎩⎪⎨

⎪⎧ , , > 0; + + =

− = − = − == ; = ; =

= = = ; = =

⇔ = = = .

Maximum of be: then = = = .

90. , , … , > 0, ∈ ℕ∗,∑ =

Find:

= ( )

Proposed by Madan Beniwal-Varanasi-India


Let = ∏ ( ) ⇒ = ∑ ( )

Let = ∑ ( ) + ∑ − ⇒ = + ( = , , … , )

Set = ⇒ − = ( = , , … , ). Thus,

= = ⋯ = = − ⇒ = − , = − , … , = − . Now,

= = (− ) ( ) ⇒ − = ( + ) = +

Thus, = − =

=+

⋯

=+

( )

www.ssmrmh.ro

91. If , , are maximum positives values such that:

( ) = − −

find: + + +

Proposed by Shivam Sharma-New Delhi-India

Solution 1 by Khalef Ruhemi-Iordania, Solution 2 by Mohammed Hijazi-

Iordania

Solution 1 by Khalef Ruhemi-Iordania

Find ∑ ( ), use ∫ ⋅ = ( ) + ; : Euler’s constant

∴ ≔ ( ) = − +−−

= − +−− ⋅ = − + − −

use ∑ = ,∑ =

Then = − + ∫ − ( )( )

∴ = − + ∫ + = − + ∫( )

( )( )

∴ = − +( − + − )

( − )

→

−−

( − )

Since → = → = =

Then, = − + − + ⋅ ∫

www.ssmrmh.ro

= − + − +( )( ) + = − + − + + ( )

∴ = − + ( ), But ( ) = − +

= − + (− + ) = − − +

∴ ∑ ( ) = − − ( − ). But = −

∴ ( ) = − − − + = − − − +

= − − ∴ ∑ ( ) = − − . Take =

⇛ ( ) = ( ) − ( ) − ( ) = − −

∴ = = = = ∴ + + + = ( )( ) =

Solution 2 by Mohammed Hijazi-Iordania

( ) = ( − ) = − +

using summation by parts:

⋅ = −+

= ( + ) − ( + )

so the red sum = ( ) − ( )

so ∑ ( ) = + −

so by comparing the result in the problem we get:

= = and = = − so + − − = ( ) =

92. Let , , be positive real numbers. Find the minimum possible value of

www.ssmrmh.ro

++

++ +

+


Solution 1 by Kevin Soto Palacios – Huarmey – Peru , Solution 2 Seyran

Ibrahimov-Maasilli-Azerbaidian , Solution 3 Imad Zak-Saida-Lebanon ,

Solution 4 Myagmarsuren Yadamsuren-Darkhan-Mongolia ,

Solution 5 Soumava Chakraborty-Kolkata-India


Siendo , , números . Hallar el mínimo valor + + +

( )( ) + ( ) ( )( ) ≥

( ) ( )

( )( ) + ( ) ( )( )

( )( ) + ( ) ( )( ) ≥

( ) ( ) ( )( )( ) + ( ) ( )

( )

Luego, aplicando MA ≥ MG

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

( ) + ( ) ( )( ) ≥

≥( + ) ( + ) + ( + )

( + )( + ) ⋅( + ) + ( + )

( + )

≥( ) ( )

( )( ) ≥ . Por transitividad + + + ≥

La igualdad se alcanza cuando = = .

Solution 2 Seyran Ibrahimov-Maasilli-Azerbaidian

+ + ≥ Nesbit = = =

www.ssmrmh.ro

+ + + ≥ − + + ; ( ) = − + +

( ) = − + ; ( ) = ⇒ − = ⇒ + = ⇒ =

( ) > 0 ⇒ = − + + = ; =

Solution 3 Imad Zak-Saida-Lebanon

= + + + ≥? ? homogeneous ⇒ + + =

= + < 3 = . By AM-GM ≤

Replace by − ; + by − + by − to get!

≥ − +−

− ++

−= − + − +

−

Note that ≥ ( − ) ( ) ⇔ ( − )( − ) ≥ True

Use AM – GM again to get

≥ − + − + − + − +−

+−

≥ − + ( − ) ( ) ≥ ≥ − + =

∴ ≥ << = >> when = = ≤

⇒ = = ⇒ = . In general = = .

Solution 4 Myagmarsuren Yadamsuren-Darkhan-Mongolia

www.ssmrmh.ro

++

++ +

+≥

1) ⋅ ⋅ ( )( ) + = ( ) ( )

( )

2) + ≥ ( )( ) ( ) (ASSURE)

≥ ≥ ; ≥

+ + + ≥ ⋅ ( + ) ⋅ + + +

≥ ( + ) ⋅ ( + ) + ( + ) =( + )

( + ) + ( + )

3) → LHS ⇒ ≥ ⋅( )( ) ( ) + ⋅ ( ) ( )

( ) =

=( + )

( + ) + ( + ) +( + ) + ( + )

( + ) +( + ) + ( + )

( + ) ≥

Solution 5 Soumava Chakraborty-Kolkata-India

If , , > 0, then, + + + =?

+ + + =( + ) + ( + )

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

( + )( + )

≥( + ) + ( + )

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

( + )( + )

≥( + ) ( + )( + )

( + )( + ) =+

( + )( + )

www.ssmrmh.ro

≥( )

( ) ∴ ≥ ( ) + ( ) (using (1))

= ( ) + ( ) + ( ) ≥ ( )( )( )( ) =

∴ required min value =

93. , , ∈ ( ,∞)

( , , ) =+ ⋅ +

+

Find ( , , )


Solution 1 by Subhajit Chattopadhyay-Visva Bharati-India , Solution 2 by

Kevin Soto Palacios – Huarmey – Peru , Solution 3 by Geanina Tudose –

Romania

Solution 1 by Subhajit Chattopadhyay-Visva Bharati-India

( , , ) =+ ⋅ +

+

take = , = ; = clearly =

Hence the LHS turns out to be

∑ = ∑ ( ) = ( + + ) − + +

Now HM ≤ AM ⇒ − ≥ − ⇒ − ≥ − ⇒ − ≥ −

∴ = ≥ ( + + ) − ( + + ) = ( + + ) ≥ =

Hence, ( ) = . Equality occurs at = = = or, = = =


www.ssmrmh.ro

Si , , ∈< 1,∞ > de tal manera que ( , , ) = ∑ ⋅

Hallar ( , , ). Realizamos los siguientes cambios de variables

= > 0, = > 0, = > 0 ⇔ =

La desigualdad propuesta es equivalente

+ ++

≥( + )( + ) = ( + ) = ( + + ) ≥

≥ = (Válidor por ≥ )

La igualdad se alcanza cuando = =

Solution 3 by Geanina Tudose – Romania

We can denote = ; = ; =

( , , ) = ( , , ) =+ +

+

Where , , > 0 subject to =

We have ( , , ) = ∑ ( ) = ∑ ( + ) −

From ≤ we have – ≥ −

⇒ ( , , ) ≥ ( + )−+

=( + )

=

= ( + + ) ≥ ⋅ =

( , , ) = ( , , ) ≥ (min. value attained for

= = = i.e. = = )

94. Let , , be positive real numbers such that: + + = . Find the

minimum of expression:

www.ssmrmh.ro

=√ +

++

+√ +

Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam

Solution by Hoang Le Nhat Tung – Hanoi – Vietnam

By inequality AM-GM. We have:

√ +=

( + )( − + )≥

⋅ + + − + = ( − + )

Similar: ≥ ; ≥ . Therefore:

⇒ = + + ≥ + + (1)

Other, by inequality CBS:

+ + = + + ≥

≥ (2)

Then (1), (2): ⇒ ≥ (3)

We will prove that: ≥ (4)

(4) ⇔ ( + + ) ≥ ( − + ) + ( − + ) + ( − + )

⇔ ( + + ) + ( + + ) ≥ ( + + ) −

−( + + ) + ( + + )

⇔ ( + + ) + ( + + ) ≥ ( + + ) −

− ( + + ) + ( + + )

⇔ ( + + ) + ( + + ) ≥

≥ ( + + )− ( + + )( + + ) +

+ ( + + ) ( + + )

(Because = + + and = ( + + ) )

⇔ ( + + ) + ( + + ) +

www.ssmrmh.ro

+ + + + + + + ( + + ) ≥

≥ ( + + ) + ( + + + + + )( + + )

⇔ ( + + ) + ( + + ) +

+ + + + + + + ( + + ) ≥

≥ ( + + ) + ( + + ) + ( + + ) +

+ ( + + )

⇔ ( + + ) + ( + + ) ≥ ( + + ) +

+ ( + + ) + ( + + ) (5)

By AM-GM I have:

+ + =+ + +

++ + +

++ + +

≥

≥ + +

⇒ + + ≥ + + ⇔ ( + + ) ≥ ( + + ) (6)

+ + =+ + +

++ + +

++ + +

≥

≥ + +

⇒ + + ≥ + + ⇔ ( + + ) ≥ ( + + ) (7)

+ + =( + )

+( + )

+( + )

≥

≥⋅

+⋅

+⋅

⇒ + + ≥ ( + + ) ⇔ + + ≥ ( + + ) (8)

Then (6), (7), (8):

⇒ ( + + ) + ( + + ) ≥ ( + + ) +

+ ( + + ) + ( + + ) ⇒ Inequality (5) True ⇒ (4) True

Then (3), (4): ⇒ ≥ ⇒ = . Equality occurs if:

www.ssmrmh.ro

⇔

⎩⎪⎨

⎪⎧

, , > 0; + + = 3+ = − + ; + = − + ; + = − +

( − + ) = ( − + ) = ( − + )= = > 0

Therefore Minimum of is: then = = = .

95. Let , , be non-negative real numbers such that + + = . Find the

maximum and minimum possible values of

( + )√ + + ( + ) + + ( + )√ + .



Siendo , , números reales no negativos de tal manera que

+ + = . Hallar el máximo y mínimo valor de

= ( + ) ( + ) + ( + ) ( + ) + ( + )√ +

Para hallar el máximo valor. Aplicamos la desigualdad de Cauchy

= ( + ) ( + ) + ( + ) ( + ) + ( + )√ + ≤

≤ ( + + + + ) ( + )( + ) + ( + )( + ) + ( + )( + )

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

= + ( + + ) ≤ +( + + )

=

= + = =√

La igualdad se alcanza cuando = = =

www.ssmrmh.ro

Para hallar el mínimo valor

Como , , ≥ ⇔ √ + ≥ , + ≥ ,√ + ≥

⇒ = ( + ) ( + ) + ( + ) ( + ) + ( + )√ + ≥

≥ ( + ) + ( + ) + ( + ) = ( + + ) =

La igualdad se alcanza cuando = , = = y sus permutaciones.

96. If , ∈ (ℝ), = then:

( + + )( + + )( + + ) ≥



Let =×

where ∈ ℂ, =×

( ) = ( ) ( ) = ( ) ( ) = | ( )| ≥

Now, + + = ( + )( + ) = ( + )( + )

where = √ ∴ ( + + ) = | ( + )| ≥

Similarly ( + + ) ≥ and ( + + ) ≥

∴ {( + + )( + + )( + + )} ≥

97. If , , ≥ , + + = then:

+ + + + + ++ + + + + ++ + + + + +

≤ ( − ) ( − ) ( − )




www.ssmrmh.ro


= +

[All the other vanishes after splitting]

= ( − )( − )( − )∑ − ( − )( − )( − ) ∑

= ( − )( − )( − ) ( − )

= ( − ) ( − ) ( − )

Now, ≤ = As + + =

∴ ≤ ( − ) ( − ) ( − )


=+ + + + + ++ + + + + ++ + + + + +

=

= ⋅ = ( ) {( − ) ( − ) ( − ) }

But + + = , , , ≥ ⇒ ( ) ≤ ( + + ) = ⇒ ( ) ≤

∴ ≤ ( − ) ( − ) ( − )

98. If , , ∈ (ℝ), = = = = = = then:

www.ssmrmh.ro

( + + + + + + ) ≥



Let =×

where ∈ ℂ and =×

, then

( ) = ( ) ( ) = ( ) ( ) = | ( )| ≥

Assuming , , commute so that = = = = = =

Now, le = + + ⇒ = ( + + ) = + + t

Let = + + + + + + = + +

= ( − )( − ), = − + √ = ( − )( − )

Now, ( ) = | ( − )| ≥

99. If , , ≥ then:

+ ++ ++ +

≤ ( + )





−−

−=

www.ssmrmh.ro

= −−

−

= −−

=

= ⋅ −−

= ≤ ( + ) ( + ) ( + )

⇒ ≤ ( + )( + )( + ); + ≥ √ ; + ≥ √ ;

+ ≥ √

⇒ ( + )( + )( + ) ≥


+ ++ ++ +

=−

−−

− → , = , ,

= −−

−

= −−

−

= −−

−

− → , = , ,

= −−

= = ( )( )( )

≤ ( + ) ( + ) ( + ) , ( + ) − ( − ) = ⇒ ( + ) ≥

= ( + )

100. If , , ∈ [ , ] then:

www.ssmrmh.ro

+ + +

≤



Put = , = , = ; ≤ , , ≤

The given determinant becomes

=

= ( )( − ) + ( − )

+ ( − )

As ( − ) + ( − ) + ( − ) =

Either two of them non-negative and one is non-positive or one of them

is non-negative and two are non-positive.

Case 1 − ≥ , − ≥ , − ≤

then ≤ ( − ) + ( − )

≤ ( − + − ) ∵ ≤ , ≤

⇒ ≤ ( − ) ≤ ∵ ≤ , ≤

Similarly for other such cases.

Case 2 − ≥ , − ≤ , − ≤

∴ ≤ ( − ) ≤

Similarly for other such cases.

www.ssmrmh.ro

Its nice to be important but more important its to be nice.

At this paper works a TEAM.

This is RMM TEAM.

To be continued!

Daniel Sitaru

RMM INEQUALITIES MARATHON 1 – 100 · Shivam Sharma - New Delhi – India . Solutions by Daniel...

Documents

Transcript of RMM INEQUALITIES MARATHON 1 – 100 · Shivam Sharma - New Delhi – India . Solutions by Daniel...