<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] We need to prove that
$$\sum_{cyc}\frac{x^3}{3-x^2}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\left(\frac{x^3}{3-x^2}-\frac{1}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)(2x^2+3x+3)}{3-x^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{(x-1)(2x^2+3x+3)}{3-x^2}-2(x^2-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)^2(2x^2+6x+3)}{3-x^2}\geq0.$$
Done!
</s>