\(2=x+y+z\ge2\sqrt{x\left(y+z\right)}\Rightarrow x\left(y+z\right)\le1\)
\(P=\frac{1}{x}\left(\frac{y+z}{yz}\right)\ge\frac{1}{x}\left(\frac{y+z}{\frac{1}{4}\left(y+z\right)^2}\right)=\frac{4}{x\left(y+z\right)}\ge4\)
\(P_{min}=4\) khi \(\left\{{}\begin{matrix}x=1\\y=z=\frac{1}{2}\end{matrix}\right.\)