Định lý SID-CID

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Tác giả: Phạm Kim Hùng
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 $x_{1},x_{2},...,x_{n} such that \deg f\le 3 $. Prove that the inequality $F(x_{1},x_{2},...,x_{n})\ge 0 $ holds if and only if

$F(1,0,0,...,0) \ge 0 \ ; \ F(1,1,0,0,...,0) \ge 0 \ ; \ F(1,1,1,...,1,0) \ge 0 \ ; \ F(1,1,1,...,1) \ge 0 \ ; $

Proof.
We denote $t=\frac{x_{1}+x_{2}}{2},x=x_{1},y=x_{2} $ and

$F=a\sum_{i=1}^{n}x_{i}^{3}+b\sum_{i<j}^{n}x_{i}x_{ j}(x_{i}+x_{j})+c\sum_{i<j<k}x_{i}x_{j}x_{k}\ ; $

Denote

$A=\sum_{i=3}^{n}x_{j}\ ; \ B=\sum_{i=3}^{n}x_{j}^{2}\ ; C=\sum_{2<i<j}x_{i}x_{j}\ ; $

We have

$F(x_{1},x_{2},...,x_{n})-F(2t,0,x_{3},...,x_{n}) $


$=a\left(x^{3}+y^{3}-(x+y)^{3}\right)+bxy(x+y)+b\left(x^{2}+y^{2}-(x+y)^{2}\right)A+xyA $


$=xy\left(-3a(x+y)+b(x+y)-2bA+A\right) \ ; $

and, alternatively, we have

$F(x_{1},x_{2},...,x_{n})-F(t,t,x_{3},...,x_{n}) $


$=a\left(x^{3}+y^{3}-\frac{(x+y)^{3}}{4}\right)+b(x+y)\left(xy-\frac{(x+y)^{2}}{4}\right) $


$+b\left(x^{2}+y^{2}-\frac{(x+y)^{2}}{2}\right)A+A\left(xy-\frac{(x+y)^{2}}{4}\right) $


$=\frac{(x-y)^{2}}{4}\left(3a(x+y)+-b(x+y)+2bA-A\right) \ ; $

According to two above relationships, we may conclude that at least one the two following inequalities hold

$F(x_{1},x_{2},...,x_{n})\ge F(2t,0,x_{3},...,x_{n}) \ \mbox{ \ or }F(x_{1},x_{2},...,x_{n})\ge F(t,t,x_{3},...,x_{n}) \ ; $

Let $t=\frac{1}{n}(x_{1}+x_{2}+...+x_{n}) $. According to UMV theorem (AC theorem), we conclude that the inequality $F(x_{1},x_{2},...,x_{n})\ge 0 $ holds if and only if

$F(t,0,0,...,0) \ge 0 \ ; \ F(t,t,0,0,...,0) \ge 0 \ ; \ F(t,t,t,...,t,0) \ge 0 \ ; \ F(t,t,t,...,t) \ge 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\sum_{i=1}^{n}x_{i}^{3}+b\sum_{i<j}x_{i}x_{j}(x _{i}+x_{j})+c\sum_{i<j<k}x_{i}x_{j}x_{k}+d\sum_{i= 1}^{n}x_{i}^{2}+e\sum_{i<j}x_{i}x_{j}+f\sum_{i=1}^ {n}x_{i}+g. $

For $t=x_{1}+x_{2}+...+x_{n} $, the inequality $F\ge 0 $ holds for all non-negative real numbers $x_{1},x_{2},...,x_{n} $ if and only if

$F\left(\frac{t}{n},0,...,0\right) \ge 0 ; \ $ $F\left(\frac{t}{n},\frac{t}{n},0,...,0\right) \ge 0\ ;\ $ $F\left(\frac{t}{n},\frac{t}{n},...,\frac{t}{n},0\r ight) \ ; \ $ $F\left(\frac{t}{n},\frac{t}{n},...,\frac{t}{n}\rig ht) \ge 0 \ ; $

Proof.
We will fix the sum $x_{1}+x_{2}+...+x_{n}=t=const $ and prove that the inequality $F\ge 0 $ holds for all $t> 0 $. Indeed, we can rewrite the expression F to (with the assumption that $x_{1}+x_{2}+...+x_{n}=t $)

$F=a\sum_{i=1}^{n}x_{i}^{3}+b\sum_{i<j}x_{i}x_{j}(x _{i}+x_{j})+c\sum_{i<j<k}x_{i}x_{j}x_{k}+ $


$+\left(\sum_{i=1}^{n}x_{i}\right)\left(\frac{d}{t} \sum_{i=1}^{n}x_{i}^{2}+\frac{e}{t}\sum_{i<j}x_{i} x_{j}\right) $


$+\frac{f}{t^{2}}\left(\sum_{i=1}^{n}x_{i}\right)^{ 2}\left(\sum_{i=1}^{n}x_{i}\right)+\frac{g}{t^{3}} \left(\sum_{i=1}^{n}x_{i}\right)^{3}. $

Notice that in this expression, t is an constant. Since this representation of F is symmetric and homogeneous, we may conclude that F\ge 0 if and only if (according to the above proposition)
$F\left(\frac{t}{n},0,...,0\right) \ge 0 ; \ F\left(\frac{t}{n},\frac{t}{n},0,...,0\right) \ge 0\ ;\ F\left(\frac{t}{n},\frac{t}{n},...,\frac{t}{n},0\r ight) \ ; \ F\left(\frac{t}{n},\frac{t}{n},...,\frac{t}{n}\rig ht) \ge 0 \ ; $

and the desired result follows.

Comment. Second volume can proved by UMV theorem, but I use another proof (as an idea of a member in Mathlinks) since I think it is original.
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