<s>[INST] 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
 [/INST] Using the Angel form of the Cauchy-Schwarz inequality: $\displaystyle \sum_{cyclic}\dfrac{a^3}{b+c} = \displaystyle \sum_{cyclic}\dfrac{(a^2)^2}{ab+ac}\ge \dfrac{(a^2+b^2+c^2)^2}{2(ab+bc+ca)}\ge \dfrac{a^2+b^2+c^2}{2}$ since $a^2+b^2+c^2 \ge ab+bc+ca$ which is clear.
</s>