How much is the minimum of $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$

We have three unknowns if they were two it were easy but I have no idea when it becomes three unknowns any hints?

note:There isn't any information about value of $a$,$b$ and $c$
Using AM-GM inequality, we have


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


The equality is indeed attained when


  $$\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\tag{1}$$


From $(1)$, we must have $a=b=c$ to attain equality.

Note:


The AM-GM inequality holds true in this case, only when $a,b,c&gt;0$ or $a,b,c&lt;0$. This can be seen by substituting $x=\frac{a}{b},y=\frac{b}{c},
z=\frac{c}{a}$ and setting $x,y,z &gt;0$
A simple counter-example to show that the sign of $a,b,c$ must be same is $a=b=-1$ and $c=1$.(This can also be proved using contradiction in the previous argument.)
If $a,b,c \in \mathbb{R}$ and no other condition is given, then as pointed out by @Behrouz Maleki, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \rightarrow -\infty.$
Well, the best I can think of, [that you stated the minimum, not the least absolute value]:


if $A$ is some arbitrarily large negative, i.e. $-10^{100}$
and $B$ is some arbitrarily small positive, i.e. $10^{-100}$ (so therefore $A/B= -10^{200}$)
and $C$ is some arbitrarily non-zero number


As $B$ decreases, $B/C$ approaches $0$, and as $A$ approaches $-\infty$,  $C/A$ approaches $0$ as well, but $A/B$ approaches $-\infty$.

As for minimizing the absolute value, that would take much more complex [or just different] maths. To summarize, $A=-\infty$ and $B=1/\infty$ and $C\ne0$

I am almost certain you meant minimizing the absolute value... but just in case, here's what I could think of.