Solution with calculus. Let $t=a+b+c$. Then $1<t\le\sqrt{3}$, and the given quantity is $\phi(t):=t+6/(t^2-1)$. Taking derivatives, we get $\phi'(t) = 1-3/(t-1)^2+3/(t+1)^2$. It is not hard to verify $\phi'<0$ on $(1,\sqrt{3})$ (in fact it is negative on a much broader interval, see here.). As a result, $\phi$ is minimized when $t=\sqrt{3}$, yielding the value $3+\sqrt{3}$.