đặt $\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c \Rightarrow abc=1$ khi đó P=$\frac{a^{3}}{(b+1)(c+1)} +\frac{b^{3}}{(a+1)(c+1)}+\frac{c^{3}}{(a+1)(b+1)}$
ad BĐT cosi ta có $\frac{a^{3}}{(b+1)(c+1)}+\frac{1+b}{8}+\frac{1+c}{8}\geq \frac{3a}{4}$
TT $\Rightarrow P\geq \frac{a+b+c}{2}-\frac{3}{4}\geq \frac{3}{4}$ ad cosi cho $a+b+c$
dấu "=" $\Leftrightarrow x=y=z=1$