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 $a, b, c$.

Because$:$ $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} -3=\frac{ab^2 +bc^2 +ca^2 -3abc}{abc}$$
It's enough to prove$:$ $$ab^2 +bc^2 +ca^2 -3abc \geqslant 0$$ by SOS.
Let $$\text{P}=2(a+b+c)\cdot (ab^2 +bc^2 +ca^2 -3abc)$$
We have$:$
$$\text{P}=a \left( a+2\,b \right)  \left( b-c \right) ^{2}+c \left( c+2\,a
 \right)  \left( a-b \right) ^{2}+b \left( b+2\,c \right)  \left( c-a
 \right) ^{2}$$
Let me explain the method$,$ let
$$\text{P}_{\text{sos}}=\sum \left( {\it QQ}_{{1}}{a}^{2}+{\it QQ}_{{4}}ab+{\it QQ}_{{5}}ac+{\it 
QQ}_{{2}}{b}^{2}+{\it QQ}_{{6}}bc+{\it QQ}_{{3}}{c}^{2} \right) 
 \left( a-b \right) ^{2}
$$
Get an identity give $$\left\{\begin{matrix}
 -{\it QQ}_{{1}}-{\it QQ}_{{2}}=0&\\2\,{\it QQ}_{{1}}-{\it QQ}_{
{4}}-{\it QQ}_{{6}}=0&\\2\,{\it QQ}_{{2}}-{\it QQ}_{{4}}-{\it QQ}_{{5}}+
2=0&\\2\,{\it QQ}_{{3}}+{\it QQ}_{{5}}+{\it QQ}_{{6}}-4=0&\\-{\it QQ}_{{1}
}-{\it QQ}_{{2}}-2\,{\it QQ}_{{3}}+2\,{\it QQ}_{{4}}+2=0  
 &
\end{matrix}\right.$$
Solve this with $Q_i \geqslant 0 ,(i=1..6)$ give us$:$ $$ \left\{ {\it QQ}_{{1}}=0,{\it QQ}_{{2}}=0,{\it QQ}_{{3}}=1,{\it QQ}_{
{4}}=0,{\it QQ}_{{5}}=2,{\it QQ}_{{6}}=0 \right\} $$
Take it into $\text{P}_{\text{sos}}$ give the SOS form.
By the same way$,$ we have$:$ $$\text{P}=\frac{1}{3} \sum (2ab-bc-ca)^2 +\sum 2ab(b-c)^2$$
