Simple Inequalities?

Question

Prove that:

9(a^3+b^3+c^3) >= (a+b+c)^3

Where a, b and c are positive real numbers.

Answer 1

By the Power Means Inequality we have

³√((a³+b³+c³)/3) ≥ (a+b+c)/3

with equality iff a=b=c

after cubing both sides we get

(a³+b³+c³)/3 ≥ (a+b+c)³/27

27 × (a³+b³+c³)/3 ≥ (a+b+c)³

9(a³+b³+c³) ≥ (a+b+c)³

as desired