THE SUMS OF SQUARE TECHNIQUE

Võ Quốc Bá Cẩn, Phạm Thị Hằng:The sums of square technique

THE SUMS OF SQUARE TECHNIQUE

I Theorem.

Consider the following inequality

m∑a +n∑a b + p∑a b+g∑ab − m+ + +n p g ∑a bc≥

With a b c, , be real numbers.

m

m m n p pg g

>



Proof.

We rewrite the inequality as

0

cyc cyc

Note that

1

2

cyc cyc cyc

cyc cyc cyc cyc cyc

cyc cyc cyc cyc cyc

cy

a b a bc b c a bc bc a b

ca a b

∑ ∑ ∑

Then the inequality is equivalent to

1

cyc cyc

m

Moreover

1

p pg g

The inequality becomes

2 2 2 2 2

2

1

1

18

cyc

cyc

m

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 ca m

+

∑

∑

2

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 0 2 2

m

m m n p pg g

>



 then the inequality is true.

Our theorem is proved J

II Application.

Example 1 (Vasile Cirtoaje) Prove that

(a +b +c ) ≥ 3(a b+b c+c a).

Solution.

The inequality is equivalent to

cyc cyc cyc

a + a b − a b≥

From this, we get m= 1,n= 2,p= − 3,g = 0, we have

1 0

m

m m n p pg g

= >





Then using our theorem, the inequality is proved J

Example 2 (Võ Qu c Bá C n) Prove that

a +b +c + − abc a+ + ≥b c a b+b c+c a

Solution.

We have m= 1,n= 0,p= − 3,g = 0 and

1 0

m

m m n p pg g

= >





Then the inequality is proved J

Example 3 (Ph m V n Thu n) Prove that

7(a +b +c ) + 10(a b+b c+c a) ≥ 0.

Solution.

We will prove the stronger result, that is

4

27

4 2 2 3 3 2

86 51 101 34

m

n

p

g

=



 = −



⇒  =



 = −



Moreover

86 0

m

m m n p pg g

= >





Then the inequality is proved J

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

3.

a a +b b +c c ≤

Solution.

On Mathlinks inequality forum, I posted the following proof:

a a +b b +c c ≥

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

2

2

2

.

yz a x zx b y xy c z

 =





 =







=



And

then, the inequality becomes

4

cyc

x

x x yz y z ≥

∑

By the Cauchy Schwarz Inequality, we get

4

cyc

x

Our lemma is proved

Now, using our lemma with note that 2 2 2

1 1 1

,

1

a b c

a b c







we get

4 2 2

Using our lemma again, we can get the result J

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

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

= = = then our inequality becomes

−

By the Cauchy Schwarz Inequality, we get

2 2

( 2 )

x y

x xy y

It suffices to show that

2

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

10 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 that

a +b +c +a b+b c+c a≥ a b+b c+c a

Problem 2 (Ph m V n Thu n, Võ Qu c Bá C n) Prove that

27

a a+b +b b+c +c c+a ≥ a+ +b c

Problem 3 (Ph m Kim Hùng) Prove that

3

a +b +c + ab+bc+ca ≥ a b+b c+c a

Võ Qu c Bá C n

Student

Can Tho University of Medicine and Pharmacy, Can Tho, Vietnam

E-mail: can_hang2007@yahoo.com