[Zequenze]

From AM-GM ;

$$\frac{a}{b}+\frac{a}{b}+\frac{b}{c} \geqslant 3\sqrt[3]{\frac{a^2}{bc}} = 3a$$ (Since $abc = 1$)

Then $$\sum_{cyc}(\frac{a}{b}+\frac{a}{b}+\frac{b}{c}) \geqslant 3\sum_{cyc} a$$

$$\therefore 3\sum_{cyc}\frac{a}{b} \geqslant 3(a+b+c)$$

Therefore $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant a+b+c$$