How to prove an inequality I didn't know where can I start because it doesn't fit to any theories or formulas.

If $a>0$, $b>0$, $c>0$ and $a+b+c=1$, prove that:
$$\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\geq \frac{1}{2}$$


Note that using AM-GM,
$$\frac{a^3}{a^2+b^2} = a - \frac{ab^2}{a^2+b^2} \geqslant a - \frac{ab^2}{2ab} = a - \frac{b}2 $$
Hence $$\sum_{cyc} \frac{a^3}{a^2+b^2} \geqslant \frac{a+b+c}2 = \frac12$$
