Question

If a, b, c > 0 prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3$

Answer 1

Using the relation $AM\ge GM$ we have
$\frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{3}\ge {(\frac{a}{b}.\frac{b}{c}.\frac{c}{a})}^{\frac{1}{3}}$
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3$