$\frac{1}{5x^2+7y^2} \le\frac{1}{2(x+y)(x+2y)}$
$\frac{1}{5y^2+7x^2} \le \frac{1}{2(x+y)(2x+y)}$ (tự bdtd ra nhé)
$\Rightarrow P \le \frac{1}{2(x+y)}\left( {\frac{1}{x+2y}+\frac{1}{y+2x}} \right)$
$\le \frac 18(\frac 1x+\frac 1y).\frac 19(\frac 1x+\frac 1y+\frac 1y+\frac1y+\frac 1x+\frac 1x)$
$=\frac 1{24}.(\frac 1x+\frac 1y)^2 \le \frac 1{24}$