Lời giải:
\(P=\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}=\frac{x^2y^2+y^2z^2+z^2x^2}{xyz}\)
Xét
\((x^2y^2+y^2z^2+z^2x^2)^2=x^4y^4+y^4z^4+z^4x^4+2x^2y^2z^2(x^2+y^2+z^2)\) (1)
Áp dụng BĐT AM-GM:
\(x^4y^4+y^4z^4\geq 2x^2y^4z^2\)
\(y^4z^4+z^4x^4\geq 2x^2y^2z^4\)
\(x^4y^4+z^4x^4\geq 2x^4y^2z^2\)
Cộng theo vế: \(\Rightarrow 2(x^4y^4+y^4z^4+z^4x^4)\geq 2x^2y^2z^2(x^2+y^2+z^2)\)
\(\Leftrightarrow x^4y^4+y^4z^4+z^4x^4\geq x^2y^2z^2(x^2+y^2+z^2)\) (2)
Từ \((1);(2)\Rightarrow (x^2y^2+y^2z^2+z^2x^2)^2\geq 3x^2y^2z^2(x^2+y^2+z^2)\)
\(\Leftrightarrow (x^2y^2+y^2z^2+z^2x^2)^2\geq 6048x^2y^2z^2\)
\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2\geq 12\sqrt{42}xyz\)
\(\Leftrightarrow P=\frac{x^2y^2+y^2z^2+z^2x^2}{xyz}\geq 12\sqrt{42}\)
Vậy \(P_{\min}=12\sqrt{42}\Leftrightarrow x=y=z=4\sqrt{42}\)