Problem
ALG-B2-M02-P005 Sum of reciprocals
#5
★★★☆☆ Level 3 of 5
If \(a,b,c>0\) and \(a+b+c=1\), prove \(\frac1a+\frac1b+\frac1c\ge9\).
Hint 1. First find an upper bound for \(abc\).
Hint 2. Use \((a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge9\).
Using the standard AM-GM/Cauchy consequence \((a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge(1+1+1)^2=9\), and \(a+b+c=1\), we get the result.
AM-GM module training problem. Method tags: am-gm, fractions.