Problem
ALG-B2-M11-P010 Two factors in the denominator
#10
★★★★☆ Level 4 of 5
Prove for \(a,b,c>0\): \[\sum_{\mathrm{cyc}}\frac{a^2}{(a+b)(a+c)}\ge\frac34.\]
Hint. After Cauchy, compare \(a^2+b^2+c^2\) and \(ab+bc+ca\).
By Cauchy, \[\sum\frac{a^2}{(a+b)(a+c)}\ge\frac{(a+b+c)^2}{\sum(a+b)(a+c)}.\] The denominator is \(a^2+b^2+c^2+3(ab+bc+ca)\). Thus it remains to check \[4(a+b+c)^2\ge3(a^2+b^2+c^2+3ab+3bc+3ca),\] which is equivalent to \(a^2+b^2+c^2\ge ab+bc+ca\).
Teaching goal: the student should first recognise the type of estimate, then choose the tool.