Ta có$\frac{1}{a^{3}}+\frac{1}{b^{3}}+1\geq 3\sqrt[3]{\frac{1}{a^{3}b^{3}}}=\frac{3}{ab}=\frac{3c}{abc}$
Tương tự
$\frac{1}{a^{3}}+\frac{1}{c^{3}}+1\geq \frac{3b}{abc}$
$\frac{1}{b^{3}}+\frac{1}{c^3}+1\geq \frac{3a}{abc}$
$\Rightarrow\frac{2}{a^{3}}+\frac{2}{b^{3}}+\frac{2}{c^{3}}+3\geq \frac{3(a+b+c)}{abc}=\frac{9abc}{abc}=9$
$\Rightarrow \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\geq 3\Rightarrow đpcm$
Dấu bằng xảy ra$\Leftrightarrow$a=b=c=1