I am trying to prove the inequality,
$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a+b+c$$
if $a, b, c$ are positive reals and $abc=1$
I have been able to tackle a special case using the AM-GM inequality. Dividing the RHS by $\sqrt[3]{abc}$ and multiplying by $3$ we get the equivalent inequality:
$$3\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right) \geq 3\sqrt[3]{\frac{a^2}{bc}}+3\sqrt[3]{\frac{b^2}{ac}}+3\sqrt[3]{\frac{c^2}{ab}} \tag1$$
If we assume that $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{b}{a} + \frac{c}{b} + \frac{a}{c}\tag2$$
we can reduce $(1)$ to:
$$\left(\sum_{cyc}\frac{a}{b}+\frac{a}{c}+1\right)\geq 3\sqrt[3]{\frac{a^2}{bc}}+3\sqrt[3]{\frac{b^2}{ac}}+3\sqrt[3]{\frac{c^2}{ab}}$$
which is obtained by using $(2)$ and applying the AM-GM inequality once. The last inequality follows from the AM-GM inequality.
I am having trouble with the assumption $(2)$. Is there any way to adapt the method I have described above to prove the inequality without using $(2)$? If possible please include any motivation that led to your solution.
