Question

If a, b, c are the sides of a triangle ABC show that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3$.

Answer 1

Since a, b, c are the sides of a triangle we have, a , b, c > 0
Applying $AM\ge GM$ to the quantities
$a/b,b/c,c/a$
We get,
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3\sqrt[3]{3\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}$
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3$