có $\frac{x}{z}+xz\geq 2x; \frac{z}{y}+yz\geq2z$$\Rightarrow P\geq 2x-xz+2y-yz+3y=2(x+z)+y(x+y+z)-xz-yz$
=$2(x+z)+y^{2}+x(y-z)$
có $x>0; y\geq z \Rightarrow x(y-z)\geq 0$
$\Rightarrow P\geq 2(x+z)+y^{2}=2(3-y)+y^{2}=(y-1)^{2}+5\geq 5$
Dấu "=" $\Leftrightarrow x=y=z=1$