It is well-known and easily verified that
$$a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a + b + c)((a - b)^2 + (b - c)^2 + (c - a)^2). \quad (1)$$
Assume without loss of generality that $a > b > c \ge 0$. Since the numbers are integers, we obtain $a - b \ge 1$, $b - c \ge 1$ and $a - c \ge 2$.
Equation (1) now implies
$$a^3 + b^3 + c^3 - 3abc \ge \frac{1}{2}(a + b + c)(1 + 1 + 4) = 3(a + b + c)$$
as desired.
Equality holds for $a = b + 1$, $b = c + 1$ and $a = c + 2$, which are exactly the triples $(t + 2, t + 1, t)$ where $t \ge 0$ is an integer, and for all their permutations.