How to prove this inequality for $a,b,c>0$? How to prove the inequality for $a,b,c>0$ :
$$\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\geq 0$$
?
 A: 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*}
A: Using https://en.m.wikipedia.org/wiki/Rearrangement_inequality
$$\sum_{cyc}{\frac{a}{(b+c)^2}} \geq \sum_{cyc}{\frac{a}{(a+b)^2}},$$
$$\sum_{cyc}{\frac{a}{(b+c)^2}} \geq \sum_{cyc}{\frac{a}{(a+c)^2}}.$$
Summing these inequalities yields the claim. 
A: The triples $(2a-b-c,2b-a-c,2c-a-b)$ and $\left(\frac{1}{(b+c)^2},\frac{1}{(a+c)^2},\frac{1}{(a+b)^2}\right)$ are the same ordered.
Thus, by Chebyshov we obtain:
$$\sum_{cyc}\frac{2a-b-c}{(b+c)^2}\geq\frac{1}{3}\sum_{cyc}(2a-b-c)\sum_{cyc}\frac{1}{(b+c)^2}=0.$$
A: Also, we can use SOS:
$$\sum_{cyc}\frac{2a-b-c}{(b+c)^2}=\sum_{cyc}\frac{a-b-(c-a)}{(b+c)^2}=$$
$$=\sum_{cyc}(a-b)\left(\frac{1}{(b+c)^2}-\frac{1}{(a+c)^2}\right)=\sum_{cyc}\frac{(a-b)^2(a+b+2c)}{(a+c)^2(b+c)^2}\geq0.$$
