theo bất dẳng thức cô-si ta có:
\(\frac{a+b+c}{3}\ge\sqrt[3]{a\cdot b\cdot c}\)(a>0,b>0,c>0)
\(\Leftrightarrow a+b+c\ge3\sqrt[3]{a\cdot b\cdot c}\)(1)
\(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\ge\sqrt[3]{\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}}\)(a>0,b>0,c>0)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}}\)(2)
Từ (1) và (2) ta suy ra:
\(\left(a+b+c\right)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\cdot3\cdot\sqrt[3]{a\cdot b\cdot c}\cdot\sqrt[3]{\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}}\)
\(\Leftrightarrow\left(a+b+c\right)\cdot\left(\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}\right)\ge9\sqrt[3]{a\cdot b\cdot c\cdot\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}}\)
\(\Leftrightarrow\left(a+b+c\right)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (đpcm)
(chúc bạn học tốt 
)
