Áp dụng bất đẳng thức côsi cho 3 số thực dương, ta có:
\[\left\{\begin{matrix} a+b+c\ge 3\sqrt[3]{abc} & \\ \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge 3\sqrt[3]{\dfrac{1}{abc}}& \end{matrix}\right.\]
\[\to (a+b+c)\bigg(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\bigg)\ge 3\sqrt[3]{abc}. 3\sqrt[3]{\dfrac{1}{abc}}\]
\[\to (a+b+c)\bigg(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\bigg)\ge 9\]
\[\to \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge \dfrac{9}{a+b+c}\]
Dấu $"="$ xảy ra $⇔a=b=c$
