Problem
ALG-B2-M11-P015 A compound denominator
#15
★★★★★ Level 5 of 5
Prove for \(a,b,c>0\): \[\sum_{\mathrm{cyc}}\frac{a^2}{b^2+c^2+a(b+c)}\ge\frac12.\]
Hint. Find the sum of all denominators.
By Cauchy, \[\sum\frac{a^2}{b^2+c^2+a(b+c)}\ge\frac{(a+b+c)^2}{2(a^2+b^2+c^2)+2(ab+bc+ca)}=\frac{(a+b+c)^2}{2(a+b+c)^2}=\frac12.\]
Teaching goal: the student should first recognise the type of estimate, then choose the tool.