\(1\ge\frac{1}{x+1}+\frac{1}{y+2}+\frac{1}{z+3}\ge\frac{9}{x+y+z+6}\)
\(\Rightarrow x+y+z\ge3\)
\(P=\frac{x+y+z}{9}+\frac{1}{x+y+z}+\frac{8\left(x+y+z\right)}{9}\ge2\sqrt{\frac{x+y+z}{9\left(x+y+z\right)}}+\frac{8.3}{9}=\frac{10}{3}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=2\\y=1\\z=0\end{matrix}\right.\)