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.


1 Answer 1


We have $$ \frac{a}{b}+\frac{c}{a}+\frac{c}{a} \ge 3\sqrt[3]{\frac{c^2}{ab}} = 3\sqrt[3]{c^3} = 3c$$

Summing this cyclically and dividing by $3$, we are done.


Not the answer you're looking for? Browse other questions tagged .