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 $a, b, c$.

$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3 \Leftrightarrow a^2c + b^2a + c^2b \ge 3abc$$
So we use these S.O.S forms:


*

*$a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a+b+c)\sum_{cyc}(a-b)^2$.

*$a^3 + b^3 + c^3 - a^2c - b^2a - c^2b = \frac{1}{3}\sum_{cyc}a^3 + a^3 + c^3-3a^2c = \frac{1}{3}\sum_{cyc} (2a+c)(c-a)^2$.
Hence, the S.O.S form would be 
$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 = \sum_{cyc}\frac{3b+c-a}{6abc}(c-a)^2.$$

