\(S=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\geq \dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}\)
Lại có
\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}\geq \dfrac{9}{a^2+b^2+c^2+2(ab+bc+ca)}=9\)
Mặt khác
\(ab+bc+ca\leq \dfrac{1}{3}(a+b+c)^2=\dfrac{1}{3}\Rightarrow \dfrac{1}{ab+bc+ca}\geq 3\)
Suy ra \(\min S=30\) đạt tại \(a=b=c=\dfrac{1}{3}\)