Problem
ALG-B2-M05-P003 Reciprocals
#3
★★☆☆☆ Level 2 of 5
Prove for \(a,b,c>0\): \[\frac1a+\frac1b+\frac1c\ge\frac{9}{a+b+c}.\]
Hint. Use the convexity of \(f(x)=1/x\).
By Jensen, \[\frac13\left(\frac1a+\frac1b+\frac1c\right)\ge\frac{1}{(a+b+c)/3}=\frac{3}{a+b+c}.\] Multiply by \(3\).
This problem links Jensen to a familiar Cauchy-type estimate.