Ta có:
$5(x^2+y^2)\geq (2x+y)^2\Leftrightarrow \sqrt{5(x^2+y^2)}\geq 2x+y$
mà:
$x^2+y^2+9+2(xy-3x-3y)=(x+y-3)^2\geq 0\Leftrightarrow 2(x+y+xy+3)\geq 8(x+y)-(x^2+y^2+3)$
Lại có:
$6(x+1)(y+1)=(2x+2)(3y+3)\leq (\frac{2x+2+3y+3}{2})^2\leq 6^2=36$
$\rightarrow x+y+xy\leq 5$
Suy ra:
$P\geq 2(x+y+xy)-24\sqrt[3]{2(x+y+xy+3)}$
Đặt $t=x+y+xy\rightarrow t\in (0;5]$
$\rightarrow P\geq f(t)=2t-24\sqrt[3]{2t+6}$
Xét $f'(t)=.......<0$
$\rightarrow min f(t)=f(5)$
$\rightarrow .............$