Prove that $\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8$

Question

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.

Answer 1

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 $(2a-b-c)^2(5a+b+c)\geq0$.