Question Number 8991 by Chantria last updated on 11/Nov/16
$${prove} \\ $$$$\:\:\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \geqslant\mathrm{3}{abc}\:;\:\forall{a},{b},{c}\geqslant\mathrm{0} \\ $$
Answered by aydnmustafa1976 last updated on 14/Nov/16
$${A}.{M}\geqslant{G}.{M}\:\:\:\left(\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)/\mathrm{3}\geqslant\left({a}^{\mathrm{3}} .{b}^{\mathrm{3}} .{c}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{3}} \\ $$