How to prove this inequality for $a,b,c>0$? How to prove the inequality for $a,b,c>0$ :
$$\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\geq 0$$
?

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:
$$\sum_{cyc}\frac{2a-b-c}{(b+c)^2}\geq\frac{1}{3}\sum_{cyc}(2a-b-c)\sum_{cyc}\frac{1}{(b+c)^2}=0.$$
