Có $A=\frac{a}{b}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}\geq 6\sqrt[6]{\frac{1}{108}}$Dấu $"="$ có khi $\frac{a}{b}=\frac{1}{2}\sqrt{\frac{b}{c}}=\frac{1}{3}\sqrt[3]{\frac{c}{a}}=\sqrt[6]{\frac{1}{108}}$
$\Leftrightarrow a=\sqrt[6]{\frac{1}{108}}b;b=(2\sqrt[6]{\frac{1}{108}})^2c$ chẳng hạn khi $c=1,b=4\sqrt[3]{\frac{1}{108}},a=4\sqrt{\frac{1}{108}}$