<s>[INST] How to prove this inequality for $a,b,c>0$? How to prove the inequality for $a,b,c>0$ :
$$\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\geq 0$$
?
 [/INST] It's clear that the inequality 
$$
\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\ge 0
$$holds for a triple $(a,b,c)$ if and only if it holds for the triple $(ta,tb,tc)$, for any $t > 0$.

Since $a,b,c > 0$, we can assume, by an appropriate scaling, that$\;a+b+c=1$.

With that assumption, it suffices to prove the inequality
$$\frac{3a-1}{2(1-a)^2}+\frac{3b-1}{2(1-b)^2}+\frac{3c-1}{2(1-c)^2}\ge 0$$
for $a,b,c > 0\;$such that$\;a+b+c=1$.

Letting $f{\,:\,}(0,1)\to\mathbb{R}$ be given by
$$f(x)=\frac{3x-1}{2(1-x)^2}$$
we get
$$f''(x)=\frac{3(1+x)}{(1-x)^4}$$
hence, since $f''(x) > 0$ for all $x\in (0,1)$, it follows that $f\;$is convex.

Then for $a,b,c > 0\;$such that$\;a+b+c=1$, we get
\begin{align*}
&\frac{3a-1}{2(1-a)^2}+\frac{3b-1}{2(1-b)^2}+\frac{3c-1}{2(1-c)^2}\\[4pt]
=\;&f(a)+f(b)+f(c)\\[4pt]
\ge\;&3f\left(\frac{a+b+c}{3}\right)\\[4pt]
=\;&3f\left({\small{\frac{1}{3}}}\right)\\[4pt]
=\;&0
\end{align*}
</s>