TITLE: Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$
QUESTION [7 upvotes]: Let $a,b,c>0$:
Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$
My solution:
We have: $\left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y} \\\frac{1}{x}+\frac{1}{y} +\frac{1}{z} \geq \frac{9}{x+y+z}\end{matrix}\right.$
$=>S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)}$
$\leq \frac{a^2}{4}.(\frac{1}{2a+b}+\frac{1}{2a+c})+\frac{b^2}{4}.(\frac{1}{2b+c}+\frac{1}{2b+a})+\frac{c^2}{4}.(\frac{1}{2c+a}+\frac{1}{2c+b})$
$=\frac{1}{4}.[a^2.(\frac{1}{2a+b}+\frac{1}{2a+c})+b^2.(\frac{1}{2b+a}+\frac{1}{2b+c})+c^2.(\frac{1}{2c+a}+\frac{1}{2c+b})]$
$\leq \frac{1}{4}.[\frac{a^2}{9}.(\frac{2}{a}+\frac{1}{b}+\frac{2}{a}+\frac{1}{c}) +\frac{b^2}{9}.(\frac{2}{b}+\frac{1}{c}+\frac{2}{b}+\frac{1}{a}) +\frac{c^2}{9}.(\frac{2}{c}+\frac{1}{a}+\frac{2}{c}+\frac{1}{b})]$
$=\frac{1}{36}.[a^2.(\frac{4}{a}+\frac{1}{b}+\frac{1}{c})+b^2.(\frac{1}{a}+\frac{4}{b}+\frac{1}{c})+c^2.(\frac{1}{a}+\frac{1}{b}+\frac{4}{c})]$
$= \frac{1}{36}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) $  
We need prove that: $S \le \frac{1}{3}$ 
$=> S \le \frac{1}{12}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) \le 1  $  <- in there, I don't know how to do that :<
I'm trying to find a solution to continue for me or another solution, Can you help me?

REPLY [2 votes]: Another way.
$$\frac{1}{3}-\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)}=\sum_{cyc}\left(\frac{1}{9}-\frac{a^2}{(2a+b)(2a+c)}\right)=$$
$$=\frac{1}{9}\sum_{cyc}\frac{2ab+2ac+bc-5a^2}{(2a+b)(2a+c)}=\frac{1}{18}\sum_{cyc}\frac{(c-a)(5a+b)-(a-b)(5a+c)}{(2a+b)(2a+c)}=$$
$$=\frac{1}{18}\sum_{cyc}(a-b)\left(\frac{5b+c}{(2b+c)(2b+a)}-\frac{5a+c}{(2a+b)(2a+c)}\right)=$$
$$=\frac{1}{18}\sum_{cyc}\frac{(a-b)^2(c^2+10ab-ac-bc)}{(2a+b)(2a+c)(2b+a)(2b+c)}\geq$$
$$\geq\frac{1}{18}\sum_{cyc}\frac{(a-b)^2(c^2-ac-bc+ab)}{(2a+b)(2a+c)(2b+a)(2b+c)}=$$
$$=\frac{1}{18\prod\limits_{cyc}((2a+b)(2a+c))}\sum_{cyc}(a-b)^2(c-a)(c-b)(2c+a)(2c+b)=$$
$$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(b-a)(4c^2+2ac+2bc+ab)}{18\prod\limits_{cyc}((2a+b)(2a+c))}=$$
$$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(4a^2c-4a^2b+2a^2b-2a^2c+a^2c-a^2b)}{18\prod\limits_{cyc}((2a+b)(2a+c))}=$$
$$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(a^2c-a^2b)}{6\prod\limits_{cyc}((2a+b)(2a+c))}=\frac{\prod\limits_{cyc}(a-b)^2}{6\prod\limits_{cyc}((2a+b)(2a+c))}\geq0.$$