<s>[INST] Prove that $\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8$ 
Prove that $$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)}{2c^2+(a+b)^2} \le 8$$.

MY ATTEMPT:I want to make a relation between $a,b,c$. By trial I found that if we put $a=b=c=1$ then the above inequality holds(equality also holds). So by trial I assume that $a+b+c=3$. After that the three functions become of the form of the function below:
$f(x)=\frac{(x+3)^2}{2x^2+(3-x)^2}$.
I calculate the function and found that :
$f(x) \le ⅓(4x+4)$.
Am I do right . Anybody has other ideas.
 [/INST] For non-negatives $a$, $b$ and $c$ let $a+b+c=3$.
Hence,
$$8-\sum_{cyc}\frac{(2a+b+c)^2}{2a^2+(b+c)^2}=\sum_{cyc}\left(\frac{8}{3}-\frac{(a+3)^2}{2a^2+(3-a)^2}\right)=$$
$$=\frac{1}{3}\sum_{cyc}\frac{(a-1)(7a-15)}{a^2-2a+3}=\frac{1}{3}\left(\sum_{cyc}\frac{(a-1)(7a-15)}{a^2-2a+3}+4(a-1)\right)=$$
$$=\sum_{cyc}\frac{(a-1)^2(4a+3)}{3(a^2-2a+3)}\geq0$$
Also there is the following.
It's enough to prove that
$$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}\leq\frac{4(4a+b+c)}{3(a+b+c)},$$
which is $(2a-b-c)^2(5a+b+c)\geq0$.
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