Proving a well-known inequality using S.O.S Using $AM-GM$ inequality, it is easy to show for $a,b,c>0$, $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3.$$
However, I can't seem to find an S.O.S form for $a,b,c$ 
$$f(a,b,c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 = \sum_{cyc}S_A(b-c)^2 \ge  0.$$
Update:
Please note that I'm looking for an S.O.S form for $a, b, c$, or a proof that there is no S.O.S form for $a, b, c$. Substituting other variables may help to solve the problem using the S.O.S method, but those are S.O.S forms for some other variables, not $a, b, c$.

Let $c=\min\{a,b,c\}$. Hence,
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{a}{b}+\frac{b}{a}-2+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1=\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}\geq0$$
From here we can get a SOS form:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{1}{6abc}\sum_{cyc}(a-b)^2(3c+a-b)$$
