THE_SUMS_OF_SQUARE_TECHNIQUE.pdf

8/18/2019 THE_SUMS_OF_SQUARE_TECHNIQUE.pdf

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Võ Quc Bá Cn Phm Th Hng

Copyright © Vo Quoc Ba Can 1

THE SUMS OF SQUARE TECHNIQUE

I. Theorem.

Consider the following inequality4 2 2 3 3 2( ) 0

cyc cyc cyc cyc cyc

m a n a b p a b g ab m n p g a bc+ + + − + + + ≥∑ ∑ ∑ ∑ ∑With , ,a b c be real numbers.

Then this inequality holds when2 2

0

3 ( )

m

m m n p pg g

>

+ ≥ + +.

Proof.

We rewrite the inequality as

4 2 2 2 2 2 3 2

3 2

( )

0

cyc cyc cyc cyc cyc cyc

cyc cyc

m a a b m n a b a bc p a b a bc

g ab a bc

− + + − + −

+ − ≥

∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ Note that

4 2 2 2 2 2

3 2 3 2 2 2

2 2 2 2 2 2

3 2 3 2 2 2

2 2

1( )

2

( )

1 1( ) ( ) ( ) ( )( 2 )

3 3

( )

( )

cyc cyc cyc

cyc cyc cyc cyc cyc

cyc cyc cyc

cyc cyc cyc cyc cyc

cy

a a b a b

a b a bc b c a bc bc a b

bc a b ab bc ca a b a b ab ac bc

ab a bc ca ab c ca a b

ca a b

− = −

− = − = −

= − − + + + − = − + −

− = − = −

= −

∑ ∑ ∑

∑ ∑ ∑ ∑ ∑

∑ ∑ ∑

∑ ∑ ∑ ∑ ∑2 2 2 21 1( ) ( ) ( )( 2 )

3 3c cyc cyc

ab bc ca a b a b ab bc ca− + + − = − − + −∑ ∑ ∑Then the inequality is equivalent to

2 2 2 2 2

2 2 2

1( ) ( )[( ) (2 ) ( 2 ) ]

2 3

( ) 0

cyc cyc

cyc cyc

ma b a b p g ab p g bc p g ca

m n a b a bc

− + − − − + + +

+ + − ≥

∑ ∑

∑ ∑Moreover

2 2 2 2

2 21 [( ) (2 ) ( 2 ) ]

6( )cyc cyc cyca b a bc p g ab p g bc p g ca

p pg g − = − − + + +

+ +∑ ∑ ∑The inequality becomes


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2 2 2 2 2

2

2 2

2 2 2

1( ) ( )[( ) (2 ) ( 2 ) ]

2 3

[( ) (2 ) ( 2 ) ] 06( )

1[3 ( ) ( ) (2 ) ( 2 ) ]

18

cyc cyc

cyc

cyc

ma b a b p g ab p g bc p g ca

m n p g ab p g bc p g ca

p pg g

m a b p g ab p g bc p g cam

− + − − − + + +

++ − − + + + ≥

+ +

⇔ − + − − + + +

∑ ∑

∑

∑2 2

2

2 2

3 ( )[( ) (2 ) ( 2 ) ] 0

18 ( ) cyc

m m n p pg g p g ab p g bc p g ca

m p pg g

+ − − −+ − − + + + ≥

+ + ∑

From now, we can easily check that if 2 2

0

3 ( )

m

m m n p pg g

>

+ ≥ + + then the inequality is true.

Our theorem is proved. J

II. Application.

Example 1. (Vasile Cirtoaje) Prove that2 2 2 2 3 3 3( ) 3( ).a b c a b b c c a+ + ≥ + +Solution.

The inequality is equivalent to4 2 2 32 0

cyc cyc cyc

a a b a b+ − ≥∑ ∑ ∑From this, we get 1, 2, 3, 0m n p g = = = − = , we have

2 2 2 2

1 0

3 ( ) 3 1 (1 2) ( 3) ( 3) 0 0 0

m

m m n p pg g

= >

+ − − − = ⋅ ⋅ + − − − − ⋅ − =Then using our theorem, the inequality is proved. J

Example 2. (Võ Quc Bá Cn) Prove that

( )4 4 4 3 3 33 1 ( ) 3( ).a b c abc a b c a b b c c a+ + + − + + ≥ + +

Solution.

We have 1, 0, 3, 0m n p g = = = − = and

( ) ( )22 2 2

1 0

3 ( ) 3 1 (1 0) 3 3 0 0 0

m

m m n p pg g

= >

+ − − − = ⋅ ⋅ + − − − − ⋅ − =Then the inequality is proved. J

Example 3. (Phm Vn Thun) Prove that

4 4 4 3 3 37( ) 10( ) 0.a b c a b b c c a+ + + + + ≥Solution.We will prove the stronger result, that is

4

4 3 177 1027

cyc cyc cyc

a a b a+ ≥

∑ ∑ ∑


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4 2 2 3 3 286 51 101 34 102 0

86

51

101

34

cyc cyc cyc cyc cyc

a a b a b ab a bc

m

n

p

g

⇔ − + − − ≥

= = −⇒

= = −

∑ ∑ ∑ ∑ ∑

Moreover

2 2 2 2

86 0

3 ( ) 3 86 (86 51) 101 101 ( 34) ( 34) 1107 0

m

m m n p pg g

= >

+ − − − = ⋅ ⋅ − − − ⋅ − − − = >Then the inequality is proved. J

Example 4. (V ình Quý) Let , , 0, 1.a b c abc> = Prove that

2 2 2

1 1 13.

1 1 1a a b b c c+ + ≤

− + − + − +Solution.On Mathlinks inequality forum, I posted the following proof:

Lemma. If , , 0, 1,a b c abc> = then2 2 2

1 1 11.

1 1 1a a b b c c+ + ≥

+ + + + + +

Proof. From the given condition , , 0, 1a b c abc> = , there exist , , 0 x y z > such that

2

2

2

.

z a

x

zxb

xyc

= =

=

And

then, the inequality becomes4

4 2 2 21

cyc

x

x x yz y z ≥

+ +∑By the Cauchy Schwarz Inequality, we get

2 2 2

2 2 2

4

4 2 2 2 4 2 2 2 4 2 2 2 4 2 21

( ) 2

cyc cyc cyc

cyc

cyc cyc cyc cyc cyc cyc

x x x x

x x yz y z x x yz y z x y z x yz x y z

≥ = ≥ =

+ + + + + + +

∑ ∑ ∑∑ ∑ ∑ ∑ ∑ ∑ ∑

Our lemma is proved.

Now, using our lemma with note that2 2 2

2 2 2

1 1 1, , 0

,1 1 1

1

a b c

a b c

> ⋅ ⋅ =

we get


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4 2 2

4 2 4 2 4 2

2 2

2 2 2 2

1 2( 1)1 2 4

1 1 1

( 1) ( 1) 1 14 4

( 1)( 1) 1 1

cyc cyc cyc

cyc cyc cyc

x x x

x x x x x x

x x x x

x x x x x x x x

+ +≥ ⇔ ≤ ⇔ ≤

+ + + + + +

+ + + − +⇔ ≤ ⇔ + ≤

+ + − + − + + +

∑ ∑ ∑

∑ ∑ ∑Using our lemma again, we can get the result. J

Now, I will present another proof of mine based on this theorem

Since , , 0, 1,a b c abc> = there exists , , 0 x y z > such that , , y z x

a b c x y z

= = = then our inequality

becomes2 2 2 2

2 2 2 2 2 2 2 2

3 3 ( 2 )3 9 4 3 3

cyc cyc cyc cyc

x x x x y

x xy y x xy y x xy y x xy y

− ≤ ⇔ ≤ ⇔ − ≥ ⇔ ≥ − + − + − + − + ∑ ∑ ∑ ∑

By the Cauchy Schwarz Inequality, we get2

2

2 2 2 22 2

( 2 )( 2 ) ( ) ( 2 )

cyc cyc cyc

x y x y x xy y x y

x xy y

− − − + ≥ − − + ∑ ∑ ∑

It suffices to show that2

2 2 2 2

4 2 2 3 3 2

( 2 ) 3 ( 2 ) ( )

10 39 25 16 8 0

cyc cyc

cyc cyc cyc cyc cyc

x y x y x xy y

x x y x y xy x yz

− ≥ − − +

⇔ + − − − ≥

∑ ∑

∑ ∑ ∑ ∑ ∑From this, we get 10, 39, 25, 16m n p g = = = − − and

2 2 2 2

10 0

3 ( ) 3 10 (10 39) ( 25) ( 25) ( 16) ( 16) 189 0

m

m m n p pg g

= >

+ − − − = ⋅ ⋅ + − − − − ⋅ − − − = >Then using our theorem, the inequality is proved. J

III. Some problems for own study.

Problem 1. (Vasile Cirtoaje) Prove that4 4 4 3 3 3 3 3 32( ).a b c a b b c c a a b b c c a+ + + + + ≥ + +

Problem 2. (Phm Vn Thun, Võ Quc Bá Cn) Prove that

3 3 3 48( ) ( ) ( ) ( ) .27

a a b b b c c c a a b c+ + + + + ≥ + +

Problem 3. (Phm Kim Hùng) Prove that

4 4 4 2 3 3 31( ) 2( ).3

a b c ab bc ca a b b c c a+ + + + + ≥ + +

Võ Quc Bá CnStudent

Can Tho University of Medicine and Pharmacy, Can Tho, VietnamE-mail: [email protected]

mailto:[email protected]:[email protected]


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