Problem
ALG-B2-M03-P003 Sum of reciprocals
#3
★★★☆☆ Level 3 of 5
If \(a,b,c>0\) and \(a+b+c=12\), prove \(\frac1a+\frac1b+\frac1c\ge\frac34\).
Hint 1. Apply Cauchy to \(1,1,1\).
Hint 2. \((1+1+1)^2\le(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\).
By Cauchy, \[9=(1+1+1)^2\le(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)=12\left(\frac1a+\frac1b+\frac1c\right).\] The result follows.
Cauchy-Schwarz module training problem. Method tags: cauchy, reciprocals.