\(P=\frac{1}{xy}+\frac{1}{yz}\ge\frac{4}{xy+yz}=\frac{4}{y\left(x+z\right)}=\frac{4}{y\left(2-y\right)}=\frac{4}{1-\left(y-1\right)^2}\)
Do \(0< y< 2\Rightarrow0< 1-\left(y-1\right)^2\le1\Rightarrow\frac{4}{1-\left(y-1\right)^2}\ge4\)
\(\Rightarrow P_{min}=4\) khi \(\left\{{}\begin{matrix}y=1\\x=z=\frac{1}{2}\end{matrix}\right.\)
Ta có : \(x+y+z=2\Rightarrow\left(x+y+z\right)^2=4\Rightarrow4\ge4\left(x+y\right)z\Rightarrow1\ge\left(x+z\right)y\)
Lại có : \(P=\frac{1}{xy}+\frac{1}{yz}=\frac{z+x}{xyz}=\frac{\left(x+z\right).1}{xyz}\ge\frac{\left(x+z\right)\left(x+z\right).y}{xyz}=\frac{\left(x+z\right)^2.y}{xyz}\ge\frac{4xzy}{xyz}=4\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+z=y\\x=z;x+y+z=2\end{matrix}\right.\) \(\Leftrightarrow y=1;x=z=\frac{1}{2}\)
Vậy ...
