Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$ 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?
 A: The inequality looks familiar. In fact, I solved the inequality a few weeks ago from an inequality handout. Here is the solution:
We have $$\begin{align}\frac{9a^2}{(2a+b)(2a+c)} &= \frac{(2a+a)^2}{2a(a+b+c)+2a^2+bc}\\ & \leq \frac{(2a)^2}{2a(a+b+c)}+\frac{a^2}{2a^2+bc} \\ &= \frac{2a}{a+b+c}+\frac{a^2}{2a^2+bc}
\end{align}$$
from Cauchy-Schwarz inequality.
Now, $$\begin{align}9\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)} &\leq 2\sum_{cyc}\frac{a}{a+b+c}+\sum_{cyc}\frac{a^2}{2a^2+bc}\\ &= 2+\sum_{cyc}\frac{a^2}{2a^2+bc}\leq 3
\end{align}$$
which is the desired result.
Here the last inequality follows because $$\sum_{cyc}\frac{a^2}{2a^2+bc}\leq1\iff \sum_{cyc}\frac{bc}{2a^2+bc}\geq1$$ is true by Cauchy-Schwarz.
A: Let $a\geq b\geq c$.
Thus, $$\frac{1}{3}-\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)}=\sum_{cyc}\left(\frac{a}{3(a+b+c)}-\frac{a^2}{(2a+b)(2a+c)}\right)=$$
$$=\frac{1}{3(a+b+c)}\sum_{cyc}\frac{a(a-b)(a-c)}{(2a+b)(2a+c)}\geq$$
$$\geq\frac{1}{3(a+b+c)}\left(\frac{a(a-b)(a-c)}{(2a+b)(2a+c)}-\frac{b(a-b)(b-c)}{(2b+a)(2b+c)}\right)\geq$$
$$\geq\frac{1}{3(a+b+c)}\left(\frac{a(a-b)\frac{a}{b}(b-c)}{(2a+b)(2a+c)}-\frac{b(a-b)(b-c)}{(2b+a)(2b+c)}\right)=$$
$$=\frac{(a-b)(b-c)}{3b(a+b+c)}\left(\frac{a^2}{(2a+b)(2a+c)}-\frac{b^2}{(2b+a)(2b+c)}\right)\geq0$$
because $$a(2b+c)\geq b(2a+c)$$ and $$a(2b+a)\geq b(2a+b).$$
A: Another way.
After full expanding we need to prove that:
$$\sum_{cyc}(2a^4b^2+2a^4c^2-4a^3b^3+5a^4bc+4a^3b^2c+4a^3c^2b-13a^2b^2c^2)\geq0,$$ which is true by Muirhead because
$(4,2,0)\succ(3,3,0),$ $(4,1,1)\succ(2,2,2)$ and $(3,2,1)\succ(2,2,2).$
Also, we can see it by AM-GM.
A: Another way.
$$\frac{1}{3}-\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)}=\sum_{cyc}\left(\frac{1}{9}-\frac{a^2}{(2a+b)(2a+c)}\right)=$$
$$=\frac{1}{9}\sum_{cyc}\frac{2ab+2ac+bc-5a^2}{(2a+b)(2a+c)}=\frac{1}{18}\sum_{cyc}\frac{(c-a)(5a+b)-(a-b)(5a+c)}{(2a+b)(2a+c)}=$$
$$=\frac{1}{18}\sum_{cyc}(a-b)\left(\frac{5b+c}{(2b+c)(2b+a)}-\frac{5a+c}{(2a+b)(2a+c)}\right)=$$
$$=\frac{1}{18}\sum_{cyc}\frac{(a-b)^2(c^2+10ab-ac-bc)}{(2a+b)(2a+c)(2b+a)(2b+c)}\geq$$
$$\geq\frac{1}{18}\sum_{cyc}\frac{(a-b)^2(c^2-ac-bc+ab)}{(2a+b)(2a+c)(2b+a)(2b+c)}=$$
$$=\frac{1}{18\prod\limits_{cyc}((2a+b)(2a+c))}\sum_{cyc}(a-b)^2(c-a)(c-b)(2c+a)(2c+b)=$$
$$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(b-a)(4c^2+2ac+2bc+ab)}{18\prod\limits_{cyc}((2a+b)(2a+c))}=$$
$$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(4a^2c-4a^2b+2a^2b-2a^2c+a^2c-a^2b)}{18\prod\limits_{cyc}((2a+b)(2a+c))}=$$
$$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(a^2c-a^2b)}{6\prod\limits_{cyc}((2a+b)(2a+c))}=\frac{\prod\limits_{cyc}(a-b)^2}{6\prod\limits_{cyc}((2a+b)(2a+c))}\geq0.$$
