Find $f(x)$ so that $\sum\limits_{cyc}a^{2}-f(x)\left(\prod_{sym}a-\prod_{sym}(1-a)\right)\geqq3(\frac{x}{2})^{2}$ .

Question

Find $f(x)$ so that $$\sum\limits_{cyc}a^{2}- f(x)\left (\prod\limits_{sym}a- \prod\limits_{sym}(1- a) \right )\geqq 3\left ( \frac{x}{2} \right )^{2}$$

OP. Given three numbers $a, b, c$ so that $\{ a, b, c \}\subseteq [0, x], abc= (x- a)(x- b)(x- c)$. Prove that $$\sum\limits_{cyc}a^{2}\geqq 3\left ( \frac{x}{2} \right )^{2} \tag{H a i D a n g e l 2 9}$$ For $x= 1\therefore f(1)= 2$ $$\sum\limits_{cyc}a^{2}- 2\left (\prod\limits_{sym}a- \prod\limits_{sym}(1- a) \right )\geqq \frac{3}{4}$$ $$\because (a+ b+ c- 2bc- 1)^{2}+ (2b- 1)^{2}c(1- c)+ \left ( c- \frac{1}{2} \right )^{2}\geqq 0$$

Answer 1

We can assume that $\{a,b,c\}\subset(0,X).$

Thus, the condition gives: $$\frac{X-a}{a}\cdot\frac{X-b}{b}\cdot\frac{X-c}{c}=1.$$ Let $\frac{X-a}{a}=\frac{x}{y}$ and $\frac{X-b}{b}=\frac{y}{z}$, where $x$, $y$ and $z$ are positives.

Thus, $\frac{X-c}{c}=\frac{z}{x}$ and we need to prove that $$\sum_{cyc}\frac{y^2}{(x+y)^2}\geq\frac{3}{4}.$$ Indeed, by C-S $$\sum_{cyc}\frac{y^2}{(x+y)^2}=\sum_{cyc}\frac{y^2(y+z)^2}{(x+y)^2(y+z)^2}\geq\frac{\left(\sum\limits_{cyc}(y^2+yz)\right)^2}{\sum\limits_{cyc}(x+y)^2(y+z)^2}.$$ Id est, it's enough to prove that $$\frac{\left(\sum\limits_{cyc}(y^2+yz)\right)^2}{\sum\limits_{cyc}(x+y)^2(y+z)^2}\geq\frac{3}{4}$$ or $$\sum_{cyc}(x^4+2x^3y+2x^3z+3x^2y^2-8x^2yz)\geq0,$$ which is true by Muirhead.

Answer 2

You're already on your way to success.

Because the expression

$$ a^2 + b^2 + c^2 \geqslant X^2 $$

is a homogeneous expression, we can easily do the following substitution:

$$ a'= \cfrac aX ,\, b'= \cfrac bX ,\, c'= \cfrac cX ,\, X'= \cfrac XX = 1 $$

where we can easily find out that:

$$ 0 \leqslant a', b', c' \leqslant X' = 1$$

Can you prove the following inequality on your own?

$$ a'^2 + b'^2 + c'^2 \geqslant \cfrac 34X'^2$$

Can you use the above proof to solve your original question?

Answer 3

Okay, let me try! After some transformations we have $$X< 2\sqrt{\frac{a^2+ b^2+ c^2}{3}}$$ Then $$abc= (X- a)(X- b)(X- c)$$ So assume that $$a= X- a,\,b= X- b,\,c= X- c$$ Then we have $$X = 2a = 2b = 2c$$ $$\therefore 4a> 2a< 2\sqrt{\frac{12a^2}{3}}$$ But this only with assumption!