Theo bđt AM-GM :
\(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xy}{z}\cdot\frac{yz}{x}}=2y\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{xy}{z}=\frac{yz}{x}\Leftrightarrow x=z\)
+ Tương tự ta cm đc :
\(\frac{yz}{x}+\frac{zx}{y}\ge2z\). Dấu "=" xảy ra <=> x = y
\(\frac{xy}{z}+\frac{xz}{y}\ge2x\). Dấu "=" xảy ra <=> y = z
Do đó : \(2P\ge2\left(x+y+z\right)\)
\(\Rightarrow P\ge1\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)
\(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xzy^2}{xz}}=2y\) ; \(\frac{xy}{z}+\frac{xz}{y}\ge2x\); \(\frac{yz}{x}+\frac{xz}{y}\ge2z\)
Cộng vế với vế:
\(2P\ge2\left(x+y+z\right)\Rightarrow P\ge x+y+z=1\)
\(\Rightarrow P_{min}=1\) khi \(x=y=z=\frac{1}{3}\)
