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Vietnam Inequality Forum

www.batdangthuc.net

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Articles Written by Member:

Pham Kim Hung

User Group: Admin

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This product is created for educational purpose Please don’t use it for any commecial purpose unless you got the right of the author Please contact www.batdangthuc.net for more details.

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Pham Kim Hung

Nhom: Admin

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Bai Viet Nay (cung voi file PDF di kem) duoc tao ra

vi muc dich giao duc Khong duoc su dung ban EBOOK nay duoi bat ky muc dich thuong mai nao, tru khi duoc su dong y cua tac gia Moi chi tiet xin

lien he: www.batdangthuc.net.

Since this following theorems are useful to solve recent problem, I decide to share them for every Mathlinkers from my book Detailed feature and more applications will be shown in this book published in Gil 2007 (IMO) Notice that this is not in Vietnamese version of 2006

Theorem 1 Suppose that F is a homogeneous symmetric polynomial of n variables

x1, x2, , xnsuch that deg f ≤ 3 Prove that the inequality F (x1, x2, , xn) ≥ 0 holds

if and only if

F (1, 0, 0, , 0) ≥ 0 ; F (1, 1, 0, 0, , 0) ≥ 0 ; F (1, 1, 1, , 1, 0) ≥ 0 ; F (1, 1, 1, , 1) ≥ 0 ;

Proof We denote t = x1 +x2

2 , x = x1, y = x2and

F = a

n

X

i=1

x3i + b

n

X

i<j

xixj(xi+ xj) + c X

i<j<k

xixjxk ; Denote

A =

n

X

i=3

xj ; B =

n

X

i=3

x2j ; C = X

2<i<j

xixj ;

We have

F (x1, x2, , xn) − F (2t, 0, x3, , xn)

= a x3+ y3− (x + y)3

+ bxy(x + y) + b x2+ y2− (x + y)2

A + xyA

= xy (−3a(x + y) + b(x + y) − 2bA + A) ;

and, alternatively, we have

F (x1, x2, , xn) − F (t, t, x3, , xn)

= a



x3+ y3−(x + y)3

4



+ b(x + y)



xy − (x + y)

2

4



+b



x2+ y2−(x + y)2

2



A + A



xy − (x + y)

2

4



=(x − y)

2

4 (3a(x + y) + −b(x + y) + 2bA − A) ;

Let t = 1

n(x1+ x2+ + xn) According to UMV theorem (AC theorem), we conclude

that the inequalityF (x1, x2, , xn) ≥ 0 holds if and only if

F (t, 0, 0, , 0) ≥ 0 ; F (t, t, 0, 0, , 0) ≥ 0 ; F (t, t, t, , t, 0) ≥ 0 ; F (t, t, t, , t) ≥ 0 ;

Since the inequality is homogeneous, we have the desired result immediately

∇

Theorem 2 (”Symmetric inequality of Degree 3” theorem - SID theorem).

Consider the following symmetric expression (not necessarily homogeneous)

F = a

n

X

i=1

x3i+ bX

i<j

xixj(xi+ xj) + c X

i<j<k

xixjxk+ d

n

X

i=1

x2i+ eX

i<j

xixj+ f

n

X

i=1

xi+ g.

For t = x1+ x2+ + xn, the inequality F ≥ 0 holds for all non-negative real numbers

x1, x2, , xnif and only if

F

t

n , 0, , 0



≥ 0; F

t

n ,

t

n , 0, , 0



≥ 0 ; F

t

n ,

t

n , ,

t

n , 0



; F

t

n ,

t

n , ,

t n



≥ 0 ;

Proof We will fix the sum x1+ x2+ + xn = t = const and prove that the inequality F ≥ 0 holds for all t > 0 Indeed, we can rewrite the expression F to (with the assumption that x1+ x2+ + xn= t)

F = a

n

X

i=1

x3i + bX

i<j

xixj(xi+ xj) + c X

i<j<k

xixjxk+

+

n

X

i=1

xi

! 

d

t

n

X

i=1

x2i +e

t

X

i<j

xixj





+f

t2

n

X

i=1

xi

!2 n

X

i=1

xi

! + g

t3

n

X

i=1

xi

!3

.

Notice that in this expression, t is an constant Since this representation of F is symmetric and homogeneous, we may conclude that F ≥ 0 if and only if (according

to the above proposition)

F



t

n , 0, , 0



≥ 0; F



t

n ,

t

n , 0, , 0



≥ 0 ; F



t

n ,

t

n , ,

t

n , 0



; F



t

n ,

t

n , ,

t n



≥ 0 ; and the desired result follows

∇