ta có $\frac{1}{x^{3}(y+z)} +\frac{y+z}{4yz}\geq 2\sqrt{\frac{y+z}{4x^{3}(y+z)yz}} =\frac{1}{x}$ do xyz=1 $\Rightarrow \frac{1}{x^{3}(y+z)}\geq \frac{1}{x}- \frac{1}{4}(\frac{1}{y}+ \frac{1}{z}$)
$\Rightarrow P \geq \frac{1}{x}+ \frac{1}{y} +\frac{1}{z} -\frac{1}{2} (\frac{1}{x}+ \frac{1}{y}+\frac{1}{z})$
$ =\frac{1}{2} (\frac{1}{x} +\frac{1}{y} +\frac{1}{z})\geq \frac{1}{2} 3\sqrt[3]{\frac{1}{x} \frac{1}{y} \frac{1}{z}}$
$=\frac{3}{2}$
dấu "=" $\Leftrightarrow x=y=z=1$