$a+\dfrac{1}{9a}+\dfrac{1}{9a}+...+\dfrac{1}{9a} \ge 10 \sqrt[10]{\dfrac{1}{(9a)^8}}$
$b+\dfrac{1}{9b}+\dfrac{1}{9b}+...+\dfrac{1}{9b} \ge 10 \sqrt[10]{\dfrac{1}{(9b)^8}}$
$c+\dfrac{1}{9c}+\dfrac{1}{9c}+...+\dfrac{1}{9c} \ge 10 \sqrt[10]{\dfrac{1}{(9c)^8}}$
$\Rightarrow (a+\dfrac{1}{a})(b+\frac{1}{b})(c+\frac{1}{c}) \ge 10^3 \sqrt[10]{\dfrac{1}{9^{27}(abc)^8}}\ge 10^3 \sqrt[10]{\dfrac{1}{9^{27}.\bigg (\dfrac{(a+b+c)^3}{27} \bigg)^8}}$
$=10^3 \sqrt[10]{\dfrac{1}{3^{30}}} =\bigg (\dfrac{10}{3} \bigg )^3$
dấu $=$ khi chỉ khi $a=b=c=\frac{1}{3}$