Inequality. $\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $ Let $x,y,z$ be real positive numbers such that $x^2+y^2+z^2=3$. Prove that : 
$$\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $$
I try to write this expression as: 
$$\frac{x^4}{x(y^2+z^2)}+\frac{y^4}{y(z^2+x^2)}+\frac{z^4}{z(x^2+y^2)}$$ and then I try to apply Cauchy-Buniakowsky but still nothing. 
I need a proof/idea without derivatives.
thanks for your help. 

By Cauchy-Schwarz you want to show that $\sum x(y^2+z^2) \leq 6$. But this is equivalent to
$$\begin{eqnarray}\sum x(3-x^2) \leq 6
\\ \Leftrightarrow 3\sum x \leq 6 + \sum x^3\end{eqnarray}$$
This then follows from $x^3 + 1 + 1 \ge 3x$ (AM-GM) and add up.
