The inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3$

Question

Suppose $a,b,c$ are positive numbers. I would like to prove the following inequality in the most elementary way:

$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 3.$$

I think I can prove using derivatives but I need to find a more simple proof.

Answer 1

This is just the inequality between the arithmetic and the geometric mean: $$\frac ab + \frac bc + \frac ca \geq 3\sqrt[3]{\frac ab \cdot \frac bc \cdot \frac ca} = 3$$