cho x+y+z=4
cmr \(\frac{1}{xy}+\frac{1}{yz}\ge1\)
BL
TA CẦN CM \(\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge1\Leftrightarrow\frac{1}{y}+\frac{1}{z}\ge x\)
mà x=\(4-\left(y+z\right)\)
\(\Rightarrow\frac{1}{y}+\frac{1}{z}\ge4-\left(y+z\right)\Leftrightarrow\frac{1}{y}-2+y+\frac{1}{z}-2+z\ge0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{y}}-\sqrt{y}\right)^2+\left(\frac{1}{\sqrt{z}}-\sqrt{z}\right)^2\ge0\)(luôn đúng)
\(\Leftrightarrow\frac{4}{x\left(y+z\right)}\ge1\)
mà \(x\left(y+z\right)\le\frac{\left(x+y+z\right)^2}{4}\)
\(\Rightarrow\frac{4}{x\left(y+z\right)}\ge\frac{4}{\frac{\left(x+y+z\right)^2}{4}}=\frac{16}{\left(x+y+z\right)^2}=\frac{16}{16}=1\left(đpcm\right)\)
