Problem
ALG-B2-M11-P008 Cube and a quadratic denominator
#8
★★★★☆ Level 4 of 5
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 \(\frac{x^3}{x^2+xy+y^2}\ge\frac{2x-y}{3}\).
For \(x,y>0\), \[3x^3-(2x-y)(x^2+xy+y^2)=(x-y)^2(x+y)\ge0.\] Hence \(\frac{x^3}{x^2+xy+y^2}\ge\frac{2x-y}{3}\). Summing for \((a,b)\), \((b,c)\), \((c,a)\), gives \(\frac{2a-b+2b-c+2c-a}{3}=\frac{a+b+c}{3}\).
Teaching goal: the student should first recognise the type of estimate, then choose the tool.