Áp Dụng BĐT cauchy schwarz ta có $\sqrt{\left ( 2x^2y^2+3x^2z^2 \right )\left ( \frac{2}{9}+\frac{3}{9} \right )}\geq \sqrt{\left ( \frac{2xy}{3}+\frac{3xz}{3} \right )^2}=\frac{2xy+3yz}{3}$
vậy ta có
$\frac{\sqrt{5}}{3}.P\geq \frac{2\sum_{}^{}xy+3\sum_{}^{} xz }{3}=\frac{5\sum_{}^{} xy}{3}=\frac{5}{3}$
$\Rightarrow P\geq \sqrt{5}$