We order the variables by size:
For $a \ge b \ge c$, all three factors are positive and we have $(a-b)(b-c)(a-c) \ge 0$.
For $b \ge c \ge a$ and $c \ge a \ge b$, two of the factors are negative and one factor is positive, so we have again $(a-b)(b-c)(a-c) \ge 0$.
For all the other orderings of variables, we have either three negative factors or one negative and two positive factors. This implies $(a-b)(b-c)(a-c) \le 0$, so the inequality holds for these cases and there is no case of equality.

Let us now consider $a \ge b \ge c$.
With the AM-GM-inequality, we get
$$(a - b)(b - c) \le \frac{(a - b + b - c)^2}{4} = \frac{(a - c)^2}{4}.$$
So we obtain
$$(a - b)(b - c)(a - c) \le \frac{(a - c)^2}{4}(a - c) = \frac{(a - c)^3}{4} \le \frac{2^3}{4} = 2.$$
The two remaining cases of orderings can be treated analogously.

We see that equality holds for $a-c=2$ and $a-b=b-c$, which implies $a=2$, $b=1$ and $c=0$. Taking into account the analogous cases, we see that equality holds exactly for the triples $(2, 1, 0)$, $(1, 0, 2)$ and $(0, 2, 1)$.

(Karl Czakler) □