Prove the given cubic inequality $a,b,c> 0$ prove $$\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$ 
I tried AM-GM but can't reached to solution please help me

Hint If you want to use AM-GM,
$$\frac{a^3}{b+c}+\frac{a(b+c)}4\ge a^2$$
Add three such inequalities to sum and then use the rearrangement $ab+bc+ca\le a^2+b^2+c^2$.
