How to prove this inequality for $a,b,c>0$? How to prove the inequality for $a,b,c>0$ :
$$\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\geq 0$$
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Also, we can use SOS:
$$\sum_{cyc}\frac{2a-b-c}{(b+c)^2}=\sum_{cyc}\frac{a-b-(c-a)}{(b+c)^2}=$$
$$=\sum_{cyc}(a-b)\left(\frac{1}{(b+c)^2}-\frac{1}{(a+c)^2}\right)=\sum_{cyc}\frac{(a-b)^2(a+b+2c)}{(a+c)^2(b+c)^2}\geq0.$$
