Problem
ALG-B1-M07-P012 Sum of reciprocals with fixed sum
#12
★★★☆☆ Level 3 of 5
Let \(x,y,z>0\) and \(x+y+z=1\). Prove that \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge9\).
Apply Cauchy to \(\left(\sum \frac{1}{x}\right)(x+y+z)\).
By Cauchy, \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+y+z)\ge(1+1+1)^2=9\). Since \(x+y+z=1\), the inequality follows. Equality holds when \(x=y=z=\frac{1}{3}\).
The student should see the paired factors \(\frac{1}{x}\) and \(x\).