Let $x,y$ and $z$ be positive numbers such that $xy+yz+zx=1$. Prove that (using Hölder's inequality):
$\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \geq \frac{(x+y+z)^3}{18}$
Thanks :)
What I try:
$\left(\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \right)\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right) \geq \left(\sum_{x,y,z}{\left(\sqrt[3]{\frac{x^3}{1+9y^2xz}\cdot\left(1+9y^2xz\right) \cdot 1}\right)}\right)^{3}=\left(\sum_{x,y,z}{x}\right)^{3}.$ So we have to prove that :
$\large\frac{\left(\sum_{x,y,z}{x}\right)^{3}}{\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right)} \geq \frac{(x+y+z)^3}{18} $ or
$3\cdot \left(3+9xyz\left( x+y+z\right)\right) \leq 18 \Leftrightarrow$ $xyz\left(x+y+z\right) \leq \frac{1}{3},$ but I don't know if this can help me to prove the inequality.
Thanks )