if a,b,c>0 and $abc=1$ prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$

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problem: a,b,c>0 and $abc=1$ prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$

my attempt:

$LHS=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=(a+b+c) \frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{a+b+c}$

From weighted AM-GM;

$LHS\geq (a+b+c) ((\frac{1}{b})^a(\frac{1}{c})^b(\frac{1}{a})^c)^{\frac{1}{a+b+c}}$

and we have :

$ ((\frac{1}{b})^a(\frac{1}{c})^b(\frac{1}{a})^c)^{\frac{1}{a+b+c}}\geq 1$

because: if $a\leq b \leq c $ :then $((\frac{1}{b})^a(\frac{1}{c})^b(\frac{1}{a})^c)\geq ((\frac{1}{b})^a(\frac{1}{c})^a(\frac{1}{a})^a)=(\frac{1}{abc})^a=1$

and if $c\leq a \leq b $ :then $((\frac{1}{b})^a(\frac{1}{c})^b(\frac{1}{a})^c)\geq ((\frac{1}{b})^c(\frac{1}{c})^c(\frac{1}{a})^c)=(\frac{1}{abc})^c=1$.and the same for others case

so finally:

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=(a+b+c) \frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{a+b+c}\geq a+b+c.$

question:

-does my attempt is true?