b2hay $x+y+1=z$ khi đó
VT=$\frac{x^{3}y^{3}}{(x+y)^{2}(x+1)^{3}(y+1)^{3}}$
$x+1\geq 3\sqrt[3]{\frac{x^{2}}{4}} \Rightarrow (x+1)^{3}\geq \frac{27}{4}x^{2}$
TT$ (y+1)^{3}\geq \frac{27}{4}y^{2}$ và có $(x+y)^{2}\geq 4xy$
$\Rightarrow (x+y)^{2}(x+1)^{3}(y+1)^{3}\geq \frac{27.27}{4}x^{3}y^{3}$
$\Rightarrow P\leq \frac{4}{729}$
dấu "=" $\Leftrightarrow x=y=2;z=5$