The maximum of $\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}$ 
Let $a,b,c$ be positive reals, such that $\frac1a+\frac1b+\frac1c=a+b+c\ (\star)$, find the maximum of $\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}$

This should be an application of Jensen's Inequality, so I have to find a concave function to maximize the sum, I thought to define;
$f(\frac1x):=\frac{1}{x+a+b+c},\quad\frac{\partial^2f(x)}{\partial x^2}<0$
so $f$ is concave and then the sum above is;
$f(\frac1a)+f(\frac1b)+f(\frac1c)$ 
using Jensen it should be less or equal to 
$3\cdot f(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}3)=3\frac{1}{\left(\frac3{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+a+b+c\right)^2}\overset{(\star)}=3\frac{1}{\left(\frac3{a+b+c}+a+b+c\right)^2}$
and the denominator can be minimized with AM-GM;
$\left(\frac3{a+b+c}+a+b+c\right)^2\ge\left(2\sqrt{3\frac{a+b+c}{a+b+c}}\right)^2=12$
so the result is $\frac14$, but it is wrong, where does it fail, maybe AM-GM value cannot be attained ?

I would rather show the maximum is $\frac3{16}$ by getting rid of the cumbersome constraint by homogenizing to:
$$\frac{abc(a+b+c)}{ab+bc+ca}\sum_{cyc} \frac1{(2a+b+c)^2} \le \frac3{16}$$
So if $a+b+c=3$, this is equivalent to 
$$16\sum_{cyc} \frac1{(3+a)^2} \le \frac1a+\frac1b+\frac1c$$
which follows from $f(a)+f(b)+f(c)\ge 0$ with
$$f(x) = \frac1x-\frac{16}{(3+x)^2}+\frac12(x-1) = \frac{(x-1)^2(x^2+7x+18)}{2x(3+x)^2} \ge 0$$
