Given $x,y,z$ is positive number that $x+y+z=1$. Prove that
$$ P = \frac{x}{y^3+2xyz} + \frac{y}{z^3+2xyz} + \frac{z}{x^3+2xyz} \gt \frac{15}{2}$$
My attempt: $$2xyz \leq \frac{2}{27}$$ $$ \Rightarrow P= \frac{x}{y^3+2xyz} + \frac{y}{z^3+2xyz} + \frac{z}{x^3+2xyz} \geq \frac{x}{y^3+\frac{2}{27}} + \frac{y}{z^3+\frac{2}{27}} + \frac{z}{x^3+\frac{2}{27}} $$ $$ \Leftrightarrow P \geq \frac{27}{2} (x+y+z) - \frac{27}{2}\left(\frac{x\cdot y^3}{y^3+\frac{2}{27}} + \frac{y\cdot z^3}{z^3+\frac{2}{27}} + \frac{z\cdot x^3}{x^3+\frac{2}{27}}\right)$$ AM-GM: $$y^3+\frac{2}{27}=y^3+\frac{1}{27}+\frac{1}{27}\geq 3\cdot\sqrt[3]{y^3\cdot\frac{1}{27}\cdot\frac{1}{27}}=\frac{y}{3}$$ Similar we have $$\frac{27}{2}\left(\frac{x\cdot y^3}{y^3+\frac{2}{27}} + \frac{y\cdot z^3}{z^3+\frac{2}{27}} + \frac{z\cdot x^3}{x^3+\frac{2}{27}}\right) \leq \frac{81}{2}\left(xy^2+yz^2+zx^2\right)$$ $$\Leftrightarrow P \geq \frac{27}{2} - \frac{81}{2}\left(xy^2+yz^2+zx^2\right)$$ Now we need prove $$\frac{81}{2}\left(xy^2+yz^2+zx^2\right)\lt \frac{27}{2} -\frac{15}{2} = 6$$ $$\Leftrightarrow xy^2+yz^2+zx^2 < \frac{4}{27}$$
I was stuck here. Help me, thank you so much.