Problem
ALG-B2-M07-P007 Cyclic ratios
#7
★★★☆☆ Level 3 of 5
Prove for \(a,b,c>0\): \[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3.\]
Hint. The product of the three fractions is \(1\).
The expression has degree \(0\). By AM-GM, \[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}}=3.\]
Connects normalization with ratios.