Question

If a, b, c $>0$, and $ab+bc+ca=6$, then prove that $abc\le 2\sqrt{2}$.

Answer 1

Given $ab+bc+ca=6$

Arthimethic mean $\ge$ Geometric mean

$⟹\frac{ab+bc+ca}{3}\ge \left(\right(ab\left)\right(bc\left)\right(ca){)}^{\frac{1}{3}}$

$⟹\frac{6}{3}\ge (abc{)}^{\frac{2}{3}}$

$⟹abc\le 2\sqrt{2}$