Problem
ALG-B2-M09-P005 Sum of reciprocals
#5
★★☆☆☆ Level 2 of 5
Let \(a,b,c>0\), \(a+b+c=5\). Prove \[\frac1a+\frac1b+\frac1c\ge\frac95.\]
Hint. Apply Cauchy to \(1,1,1\).
By Cauchy, \(9=(1+1+1)^2\le(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)=5\sum1/a\). Hence \(\sum1/a\ge9/5\).
Fixed sum and a convex function.