Ta có$\frac{a^3}{2b^3}+\frac{b^3}{2c^3}+\frac{1}{2}\geq 3\sqrt[3]{\frac{a^3b^3}{8b^33c^3}}=\frac{3}{2}\frac{a}{c}$
Tương tự
$\Rightarrow \frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\geq \frac{3}{2}(\frac{a}{c}+\frac{b}{a}+\frac{c}{b})-\frac{3}{2}\geq \frac{3}{2}.3-\frac{3}{2}=3$
Dấu bằng xảy ra khi và chỉ khi a=b=c=1