Prove that $\sum_{cyc}\frac{x}{y^2+z^2}\ge\frac{3\sqrt3}2$

Question

Given $x,y,z$ are positive number satisfy $x^2+y^2+z^2=1$. Prove that $$\frac{x}{y^2+z^2}+\frac{y}{z^2+x^2}+\frac{z}{x^2+y^2}\ge \frac{3\sqrt{3}}{2}$$

I need a way use reduction of many fractions to a common denominator

Answer 1

We need to prove that $$\sum_{cyc}\frac{x}{y^2+z^2}\geq\frac{3\sqrt3}{2\sqrt{x^2+y^2+z^2}}.$$ Since the last inequality is homogeneous, we can assume that $x^2+y^2+z^2=3$ and now we need to prove that $$\sum_{cyc}\frac{x}{y^2+z^2}\geq\frac{3}{2}$$ or $$\sum_{cyc}\left(\frac{x}{3-x^2}-\frac{1}{2}\right)\geq0$$ or $$\sum_{cyc}\left(\frac{x}{3-x^2}-\frac{1}{2}-\frac{1}{2}(x^2-1)\right)\geq0$$ or $$\sum_{cyc}\frac{x(x+2)(x-1)^2}{3-x^2}\geq0$$ Done!

Answer 2

Not hard to see $$x(1-x^2)\le\dfrac{2}{3\sqrt{3}}.$$

Then $$\dfrac{x}{1-x^2}\ge \dfrac{3\sqrt{3}}{2}x^2.$$

Then sum over $\{x,y,z\}$ and use $x^2+y^2+z^2=1$