\(S\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}+\dfrac{9}{ab+bc+ca}=a+b+c+\dfrac{9}{ab+bc+ca}\)
\(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\)
\(\Rightarrow S\ge a+b+c+\dfrac{27}{\left(a+b+c\right)^2}=a+b+c+\dfrac{1}{\left(a+b+c\right)^2}+\dfrac{26}{\left(a+b+c\right)^2}\ge\dfrac{2}{\sqrt{a+b+c}}+26\ge2+26=28\)