Prove Inequality: a,b,c>0 \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3

[deancodemo]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I'm not so sure how to do this. Usually I would try to prove that \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 \ge 0 but this gets me nowhere: \frac{a^2 c + b^2 a + c^2 b - 3abc}{abc}. I can't factorise the numerator.

I know of a similar inequality that I can prove easily using this method, which is \frac{a}{b} + \frac{b}{a} \ge 2 but the inequality above is harder for me. Please help.

 

[Dunkle]

Are a, b, and c integers?

 

[snipez90]

It doesn't matter if a, b, c are integers or not. The inequality holds for real a,b,c > 0.

Clear denominators, divide by 3, and apply the AM-GM inequality.