Đặt $\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z(x,y,z>0)$
$\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1$
Ta có: $P=\frac{x}{y^{2}}+\frac{y}{z^{2}}+\frac{z}{x^{2}}-(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}})-2$
$=\frac{x-1}{y^{2}}+\frac{y-1}{z^{2}}+\frac{z-1}{x^{2}}-2$
$=\frac{x-1+y-1}{y^{2}}+\frac{y-1+z-1}{z^{2}}+\frac{z-1+x-1}{x^{2}}-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}})-2$
$=(x-1)(\frac{1}{x^{2}}+\frac{1}{y^{2}})+(y-1)(\frac{1}{y^{2}}+\frac{1}{z^{2}})+(z-1)(\frac{1}{z^{2}}+\frac{1}{x^{2}})-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}})-2$
$\geq \frac{2(x-1)}{xy}+\frac{2(y-1)}{yz}+\frac{2(z-1)}{zx}-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}})-2$
$=(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}})-4$
Ta có$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}\geq 3(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz})=3$
$\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \sqrt{3}$
$\Rightarrow P\geq \sqrt{3}+1-4=\sqrt{3}-3$
Dấu = đạt khi a=b=c=$\frac{1}{\sqrt{3}}$
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