$\frac{6}{xy}=4+x^4+y^4 \ge 4+2x^2y^2$$\Leftrightarrow x^2y^2+2 \le \frac 3{xy}\Leftrightarrow x^3y^3+2xy-3 \le0 \Leftrightarrow xy \le 1$
$P=\frac 1{1+2x} + \frac 1{1+2y}+\frac{3-2xy}{5-x^2-y^2}$
$\ge \frac{1}{1+x^2+1} +\frac 1{1+y^2+1}+\frac{3-2.1}{5-x^2-y^2}$
$=\frac{1}{x^2+2}+ \frac{1}{y^2+2}+\frac 1{5-x^2-y^2} \ge \frac{9}{2+x^2+2+y^2+5-x^2-y^2}=1$
Vậy $\min P=1$ đạt đc khi $x=y=1$