Problem
ALG-B2-M11-P013 Holder with a cyclic denominator
#13
★★★★★ Level 5 of 5
Prove for \(a,b,c>0\): \[\sum_{\mathrm{cyc}}\frac{a^3}{b^2+bc+c^2}\ge\frac{a+b+c}{3}.\]
Hint. Use Holder: \(\sum\frac{x^3}{y}\ge\frac{(\sum x)^3}{\sum y}\).
By Holder, \[\sum\frac{a^3}{b^2+bc+c^2}\ge\frac{(a+b+c)^3}{2(a^2+b^2+c^2)+ab+bc+ca}.\] Since \(3(a+b+c)^2\ge2(a^2+b^2+c^2)+ab+bc+ca\), the lower bound is \((a+b+c)/3\).
Teaching goal: the student should first recognise the type of estimate, then choose the tool.