Problem
ALG-B1-M07-P018 Three reciprocal linear forms
#18
★★★★☆ Level 4 of 5
Let \(a,b,c>0\) and \(a+b+c=3\). Prove \[\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c}\ge\frac{3}{4}.\]
Use Cauchy: \(\sum \frac{1^2}{3+a}\ge\frac{9}{(3+a)+(3+b)+(3+c)}\).
By Cauchy, the left-hand side is at least \(\frac{(1+1+1)^2}{(3+a)+(3+b)+(3+c)}=\frac{9}{9+a+b+c}=\frac{9}{12}=\frac{3}{4}\). Equality holds when \(a=b=c=1\).
A good problem on recognizing the sum of denominators.