Problem
ALG-B2-M12-P003 Set 1. Cube in the numerator
#3
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Prove for \(a,b,c>0\): \[\sum_{\mathrm{cyc}}\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}.\]
Hint. Prove locally that \(\frac{x^3}{x^2+xy+y^2}\ge\frac{2x-y}{3}\).
We have \(3x^3-(2x-y)(x^2+xy+y^2)=(x-y)^2(x+y)\ge0\). Summing cyclically gives \(\frac{2a-b+2b-c+2c-a}{3}=\frac{a+b+c}{3}\).
Mock set 1: the problem is intended for independent method selection without an explicit cue in the statement.