Có A=ab+12√bc+12√bc+133√ca+133√ca+133√ca≥66√1108Dấu "=" 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}}