<s>[INST] 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. 
 [/INST] The function $f(w)=w^{3/2}/(3-w)$ is convex on $(0,3)$ so 
$${1\over 3}\left[f(x^2)+f(y^2)+f(z^2)\right]\geq f\left([(x^2+y^2+z^2)/3]\right).$$
</s>