Áp dụng bất đẳng thức $AM-GM$ ta được:
$1 + \dfrac{1}{3a} +\dfrac{1}{3a} +\dfrac{1}{3a}\geqslant 4\sqrt[4]{\dfrac{1}{(3a)^3}}$
$1 + \dfrac{1}{3b} +\dfrac{1}{3b} +\dfrac{1}{3b}\geqslant 4\sqrt[4]{\dfrac{1}{(3b)^3}}$
$1 + \dfrac{1}{3c} +\dfrac{1}{3c} +\dfrac{1}{3c}\geqslant 4\sqrt[4]{\dfrac{1}{(3c)^3}}$
Nhân vế theo vế ta được:
$\left(1+\dfrac1a\right)\left(1+\dfrac1b\right)\left(1+\dfrac1c\right)\geqslant 64\sqrt[4]{\dfrac{1}{(3^3.abc)^3}}$
$\Leftrightarrow \left(1+\dfrac1a\right)\left(1+\dfrac1b\right)\left(1+\dfrac1c\right)\geqslant \dfrac{64}{\sqrt[4]{(a+b+c)^9}}$
$\Leftrightarrow \left(1+\dfrac1a\right)\left(1+\dfrac1b\right)\left(1+\dfrac1c\right)\geqslant 64$
Dấu $=$ xảy ra $\Leftrightarrow a = b = c =\dfrac13$
