Question

If a, b, c $>0$, abc $=8$, then prove that $a^2+b^2+c^2\ge 12$.

Answer 1

$\frac{a^2+b^2+c^2}{3}\ge \sqrt[3]{3a^2{b}^2{c}^2}{a}^2+b^2+c^2\ge 3(8^2{)}^{1/3}{a}^2+b^2+c^2\ge 3\times 2^{\frac{6}{3}}{a}^2+b^2+c^2\ge 12$