show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $ We want to show that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq  \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ if $abc \geq 1 $
I've tried to use AM-GM directly on the LHS but that obviously failed.
Simplifying the question gives $a^2b + b^2c + c^2a \geq ab + bc + ac$
The fact that this is $a(ab) + b(bc) + c(ca) \geq ab + bc + ac$ makes me think that there's a way to proceed from here, but I'm not quite sure how.
Note : a,b,c are real positive numbers

Case 1: Let $ab+bc+ca\ge a+b+c$, then the AM-HM of $a,b,c$ with weights as $1/b,1/c,1/a$ gives
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge \frac{1/a+1/b+1/c)^2}{1/(ab)+1/(bc)+1/ca)}=\frac{(ab+bc+ca)(1/a+1/b+1/c)}{a+b+c} \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Case 2: Let $(ab+bc+ca) \le a+b+c$, then AM-HM of $1/b,1/c,1/a$ with weights as $a,b,c$, we get
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{(a+b+c)^2}{ab+bc+ca} \ge ab+bc+ca=\frac{1}{b}+\frac{1}{c}+\frac{1}{a}$$
