The discriminant of the given equation is \[ \Delta = 16(a+b+c)^2 - 48(ab+bc+ca) = 8(2a^2+2b^2+2c^2-2ab-2bc-2ca) = 8((a-b)^2 + (b-c)^2 + (c-a)^2). \]Consequently, it has two real roots and negative in the interval \[ I = \left(2(a+b+c) -\frac{\sqrt{\Delta}}{2},2(a+b+c)+\frac{\sqrt{\Delta}}{2}\right), \]whose size is larger than 1. Now if one of the endpoints of $I$ is integral, then there is a second integral number contained in $I$. Otherwise, the conclusion is clear as $|I|\ge 1$.

Furthermore, show that $\frac{1}{2\sqrt{2}}$ can be replaced by $\frac{1}{2\sqrt{3}}$ and $\frac{1}{2\sqrt{3}}$ is the best constant.

@silouan: Nice follow-up! As for your first question, here's an argument that $\frac{1}{2\sqrt{2}}$ can be replaced by $\frac{1}{2\sqrt{3}}$: Let $a\le b\le c$ wlog, and suppose $c-a>\frac{1}{2\sqrt{3}}$. Noting that $c-a=(c-b)+(b-a)$, we have by Cauchy-Schwarz that \[ 2 \left((c-b)^2 + (b-a)^2 \right)\ge (c-b+b-a)^2 >\frac{1}{12}, \]implying that \[ (a-b)^2 + (b-c)^2 + (c-a)^2 > \frac{1}{24}+\frac{1}{12} = \frac{1}{8}, \]so that $\Delta>1$. Now, if we take anything less than $1/(2\sqrt{3})$ and $b$ as the midpoint of $[a,c]$ (or a slight perturbation of it), we have $|I|<1$ for $I$ appearing in my solution. Hence, there must be examples $(a,b,c)$ for which it contains no integers (although I can't see a construction right away).

Since it is an upward opening parabola it is sufficient to show that the distance between the two roots, $r_1$ and $r_2$, is at least one: $$|r_1-r_2|=\sqrt{\left(4(a+b+c)\right)^2-4\cdot 12(ab+bc+ca)}=\sqrt{8\left((a-b)^2+(b-c)^2+(c-a)^2\right)}\geq 1$$