Prove that $$\frac{x^2}{x^2+2y}+\frac{y^2}{y^2+2z}+\frac{z^2}{z^2+2x} \ge 1$$ with $xy+yz+zx=3$ and $x,y,z >0$.
If I use Cauchy-Schwarz then $\frac{x^2}{x^2+2y}+\frac{y^2}{y^2+2z}+\frac{z^2}{z^2+2x} \ge \frac{(x+y+z)^2}{x^2+y^2+z^2+2(x+y+z)}\le 1$
How can I solve that?
Use Cauchy-Schwarz in the form $$ \frac{x^2}{x^2+2y}+\frac{y^2}{y^2+2z}+\frac{z^2}{z^2+2x}\\= \frac{x^3}{x^3+2xy}+\frac{y^3}{y^3+2yz}+\frac{z^3}{z^3+2xz} \\ \geq \dfrac{\left (\sqrt{x^3}+\sqrt{y^3}+\sqrt{z^3}\right )^2}{x^3+y^3+z^3+2(xy+yz+zx)} \\ = \dfrac{\left (\sqrt{x^3}+\sqrt{y^3}+\sqrt{z^3}\right )^2}{x^3+y^3+z^3+6} $$ So it is enough to prove $\left (\sqrt{x^3}+\sqrt{y^3}+\sqrt{z^3}\right )^2\geq x^3+y^3+z^3+6\Leftrightarrow \sqrt{x^3y^3}+\sqrt{y^3z^3}+\sqrt{z^3x^3}\geq 3$.
This can be shown using Hölder's inequality: $\left ( \sqrt{x^3y^3}+\sqrt{y^3z^3}+\sqrt{z^3x^3}\right )\left ( \sqrt{x^3y^3}+\sqrt{y^3z^3}+\sqrt{z^3x^3}\right )(1+1+1)\geq (xy+yz+zx)^3=27$ which is exactly $\sqrt{x^3y^3}+\sqrt{y^3z^3}+\sqrt{z^3x^3}\geq 3$.