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

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