$+,3=x+y+z\geq \sqrt[3]{xyz}\Rightarrow xyz\leq1$$+,P\geq 3\sqrt[3]{\frac{1}{xyz(x+1)(y+1)(z+1)}}\geq \frac{3}{\sqrt[3]{(x+1)(y+1)(z+1)}}\geq \frac{3}{\frac{x+y+z+3}{3}}=\frac{3}{2}$
$+,3=x+y+z\geq
3\sqrt[3]{xyz}\Rightarrow xyz\leq1$$+,P\geq 3\sqrt[3]{\frac{1}{xyz(x+1)(y+1)(z+1)}}\geq \frac{3}{\sqrt[3]{(x+1)(y+1)(z+1)}}\geq \frac{3}{\frac{x+y+z+3}{3}}=\frac{3}{2}$