Problem
ALG-B1-M07-P021 Product of two sums
#21
★★★★☆ Level 4 of 5
Let \(a,b,c>0\). Prove \[(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\]
This is a direct application of Cauchy-Schwarz.
By Cauchy, \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge(\sqrt{a}\cdot\frac{1}{\sqrt{a}}+\sqrt{b}\cdot\frac{1}{\sqrt{b}}+\sqrt{c}\cdot\frac{1}{\sqrt{c}})^2=9\). Equality holds when \(a=b=c\).
A classical form students should recognize quickly.