Problem
ALG-B2-M10-P013 Cyclic quadratic denominators
#13
★★★★★ Level 5 of 5
Prove for \(a,b,c>0\): \[\frac{a^2}{a^2+ab+b^2}+\frac{b^2}{b^2+bc+c^2}+\frac{c^2}{c^2+ca+a^2}\ge\frac12.\]
Hint. Apply Cauchy in Engel form, then compare the denominator with \(2(a+b+c)^2\).
By Cauchy, \[\sum_{\mathrm{cyc}}\frac{a^2}{a^2+ab+b^2}\ge\frac{(a+b+c)^2}{2(a^2+b^2+c^2)+ab+bc+ca}.\] It remains to note that \[2(a+b+c)^2-\bigl(2(a^2+b^2+c^2)+ab+bc+ca\bigr)=3(ab+bc+ca)\ge0.\] Hence the denominator is at most \(2(a+b+c)^2\), and the whole fraction is at least \(\frac12\).
Teaching goal: choose a combination of methods and always check the equality case.