$\dfrac{1}{a^2+b^2+c^2}+\sum \dfrac{1}{ab}
\\\geq \dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}
\\\geq \dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\dfrac{7}{\dfrac{(a+b+c)^2}{3}}
\\=\dfrac{9}{(a+b+c)^2}+\dfrac{21}{(a+b+c)^2}=30(dpcm)$