Problem
ALG-B2-M10-P016 Mixed fractional sum
#16
★★★★★ Level 5 of 5
Let \(a,b,c\ge0\), \(a+b+c=3\). Prove \[\sum_{\mathrm{cyc}}\frac{a^2+1}{a+b+1}\ge2.\]
Hint. Split the numerator as \(a^2+1\) and apply Cauchy to each part.
By Cauchy, \(\sum\frac{a^2}{a+b+1}\ge\frac{(a+b+c)^2}{(a+b+1)+(b+c+1)+(c+a+1)}=\frac{9}{9}=1\). Also \(\sum\frac1{a+b+1}\ge\frac{9}{9}=1\). Adding gives the result.
Teaching goal: choose a combination of methods and always check the equality case.