Statement $\frac ab+\frac bc+\frac ca \ge 3$ is true for any positive $a$, $b$, $c$?

Question

Let us have $a,b,c$ arbitrary positive numbers. Prove:

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

I have tried many things, but none of them seemed to work. Any ideas? (If duplicate, I am sorry, didn't see it yet) :)

Counterexample $a=1,b=-1,c=1\implies a/b+b/c+c/a=-1-1+1<3$. $a,b,c$ must be of the same sign, I think. — Samrat Mukhopadhyay, May 06 '15 at 07:11
$a,b,c$ of the same sign I think, consider $(a,b,c)=(1,-1,1)$. — Brian Ding, May 06 '15 at 07:11
See also: Proving $\frac{x}{y} +\frac{y}{z} + \frac{z}{x} \ge 3$ for positive $x,y,z$ and other questions linked there. — Martin Sleziak, Aug 18 '20 at 11:48

score 6 · Accepted Answer · answered May 06 '15 at 07:13

6

You can Use AM-GM inequality to get $$\frac{a/b+b/c+c/a}{3}\ge (a/b\cdot b/c\cdot c/a)^{1/3}=1$$

answered May 06 '15 at 07:13

1 Answers1