\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}x+y\ge2\sqrt{xy}\\\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\sqrt{xy.\frac{1}{xy}}\)
\(\Rightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\) ( đpcm )
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}x+y+z\ge3\sqrt{xyz}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt{\frac{1}{xyz}}\end{matrix}\right.\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\sqrt{xyz.\frac{1}{xyz}}\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) ( đpcm )
