Prove that $\sum_{cyc}\frac{(a+b-c)^2}{(a+b)^2+c^2}\ge \frac{3}{5}$ $a,b,c$ are reals $ >0$ prove that $$\sum_{cyc}\frac{(a+b-c)^2}{(a+b)^2+c^2}\ge \frac{3}{5}$$
The inequality is homogeneous so assum WLOG $a+b+c=3$, the inequality is equivalent to $$\sum_{cyc}\frac{(3-2c)^2}{(3-c)^2+c^2} \ge 3/5$$
Set $$f(x)= \frac{(3-2x)^2}{(3-x)^2+x^2} $$
We wish to prove that for $0<x<3$ we have $f(x)\ge 1/5$... but unfortunately this is false, so what should I do?

Hint: Show $f(x)+\frac{18}{25}(x-1)\geqslant \frac15$ for all $x\in (0,3)$.
