Problem
ALG-B2-M10-P001 Fractions with sum 3
#1
★★★★★ Level 5 of 5
Let \(a,b,c\ge0\), \(a+b+c=3\). Prove \[\frac{a^2}{1+a}+\frac{b^2}{1+b}+\frac{c^2}{1+c}\ge\frac32.\]
Hint. The sum of denominators is also fixed.
By Cauchy, \[\sum\frac{a^2}{1+a}\ge\frac{(a+b+c)^2}{(1+a)+(1+b)+(1+c)}=\frac{9}{6}=\frac32.\] Equality occurs at \(a=b=c=1\).
Teaching goal: choose a combination of methods and always check the equality case.