If at least one of these three numbers is zero, the relation is obvious. If none of them equals 0, then dividing the give relation by $|abc|$ gives $$\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}=0$$ Now we have three unit vectors, $\frac{a}{|a|}$, $\frac{b}{|b|}$ and $\frac{c}{|c|}$, whose sum is 0. Therefore the angles between them are all $120^\circ$ (see # 2178). Then $$|a-b| = \sqrt{a^2+b^2-2ab\cos 120^\circ} =\sqrt{a^b + b^2 +ab} = \sqrt{(a-b)^2+3ab} \ge \sqrt{3ab}$$ Hence we have $$|(a-b)(b-c)(c-a)|\ge\sqrt{3ab}\sqrt{3bc}\sqrt{3ca}=3\sqrt{3}|abc|$$