$\frac{x+y+z}{xyz}=1<=>\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1$
$a^{2}+b^{2}+c^{2}\geq ab+bc+ac=>\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\geq \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1$
$P=\frac{x-1}{y^{2}}+\frac{y-1}{z^{2}}+\frac{z-1}{x^{2}}=\frac{(x-1)+(y-1)}{y^{2}}+\frac{(y-1)+(z-1)}{z^{2}}+\frac{(z-1)+(x-1)}{x^{2}}-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}})=(x-1)(\frac{1}{y^{2}}+\frac{1}{x^{2}})+(y-1)(\frac{1}{y^{2}}+\frac{1}{z^{2}})+(z-1)(\frac{1}{x^{2}}+\frac{1}{z^{2}}) -(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}})\geq \frac{2(x-1)}{xy} +\frac{2(y-1)}{yz}+\frac{2(z-1)}{xz} -(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+1=\frac{1}{x} +\frac{1}{y}+\frac{1}{z}-2(\frac{1}{xz}+\frac{1}{yz}+\frac{1}{xy})+1$
Dễ dàng tìm được min $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$