Problem
ALG-B2-M07-P013 Sum of reciprocals
#13
★★★★☆ Level 4 of 5
Prove for \(a,b,c>0\): \[(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge9.\]
Hint. The inequality has degree \(0\). You may set \(a+b+c=1\).
By Cauchy, \((1+1+1)^2\le(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\). Therefore the product is at least \(9\).
Degree \(0\) helps reveal scale invariance.