Inequality $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b+ c$ when $abc = 1$

Solution 1:

$$a+b+c =$$

$$\{ \mbox{multiply by 1: maybe } abc\mbox{, maybe }\sqrt{abc}, ... \}$$

$$= \frac{a}{\sqrt[3]{abc}}+\frac{b}{\sqrt[3]{abc}}+\frac{c}{\sqrt[3]{abc}}=$$

$$ =\sqrt[3]{\frac{a}{b} \cdot \frac{a}{b} \cdot \frac{b}{c}} +\sqrt[3]{\frac{b}{c} \cdot \frac{b}{c} \cdot \frac{c}{a}} +\sqrt[3]{\frac{c}{a} \cdot \frac{c}{a} \cdot \frac{a}{b}} \le $$ $$ \{\mbox{ GM}\le\mbox{AM }\} $$ $$ \le \frac{\frac{a}{b}+\frac{a}{b}+\frac{b}{c}}{3} + \frac{\frac{b}{c}+\frac{b}{c}+\frac{c}{a}}{3} + \frac{\frac{c}{a}+\frac{c}{a}+\frac{a}{b}}{3} = $$ $$ =\frac{a}{b}+\frac{b}{c}+\frac{c}{a}. $$