Theo giả thiết: $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1$
Ta có:
$P=\frac{(x-1)+(y-1)}{x^2}+\frac{(y-1)+(z-1)}{y^2}+\frac{(z-1)+(x-z)}{z^2}-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
$=(x-1)(\frac{1}{x^2}+\frac{1}{z^2})+(y-1)(\frac{1}{x^2}+\frac{1}{y^2})+(z-1)(\frac{1}{y^2}+\frac{1}{z^2})-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
$\geq \frac{2(x-1)}{xz}+\frac{2(y-1)}{xy}+\frac{2(z-1)}{yz}-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
$=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-2(\frac{1}{xz}+\frac{1}{xy}+\frac{1}{yz})=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-2 (1)$
Mà $(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 \geq 3(\frac{1}{xz}+\frac{1}{xy}+\frac{1}{yz})=3 (2)$
Từ $(1)$ và $(2) \Rightarrow P \geq \sqrt{3}-2$
Kết luận: Vậy $\min P=\sqrt{3}-2 \Leftrightarrow x=y=\sqrt{3}$