$gt\Leftrightarrow (x+y)^2+z^2 =8xy+2(x+y+z) \le 2(x+y)^2+2(x+y+z)$$\Leftrightarrow (x+y)^2+2(x+y)+1 \ge z^2-2z+1$
$\Leftrightarrow (x+y+1)^2 \ge (z-1)^2\Leftrightarrow x+y+2 \ge z$
Khi đó $VT \ge \frac{x+1}{x+2y+1}+\frac{y+1}{y+2x+1}+\frac{(x+y)^2}{(x+y+2)^2} $
$\ge \frac{(x+y+2)^2}{(x+1)(x+2y+1)+(y+1)(y+2x+1)}+\frac{(x+y)^2}{(x+y+2)^2}$
$\ge \frac{(x+y+2)^2}{(x+y)^2+4(x+y)+2xy+2}+\frac{(x+y)^2}{(x+y+2)^2} \\ \ge \frac{(x+y+2)^2}{\frac 32(x+y)^2+4(x+y)+2} +\frac{(x+y)^2}{(x+y+2)^2}$
$=\frac{2(x+y+2)}{3(x+y)+1}+\frac{(x+y)^2}{(x+y+2)^2}\overset{x+y \to t \ge 2}{=}\frac{2(t+2)}{3t+2}+\frac{t^2}{(t+2)^2} \ge \frac 54$
$P_{\min}=\frac 54\Leftrightarrow x=y=1,z=4$