The desired inequality is equivalent to $a^3 + b^3 + c^3 - 3abc \ge 9k$. We have
$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).$$
Integers $|a-b|$, $|b-c|$ and $|c-a|$ can not all be equal $1$ so
$$\begin{aligned} a^2 + b^2 + c^2 - ab - bc - ca &= \frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) \\ &\ge \frac{1}{2}(1^2 + 1^2 + 2^2) = 3. \end{aligned}$$

It follows that $a^3 + b^3 + c^3 - 3abc \ge 3(a + b + c)$. Since
$$\begin{aligned} (a + b + c)^2 &= a^2 + b^2 + c^2 + 2ab + 2bc + 2ac \\ &= (a^2 + b^2 + c^2 - ab - bc - ac) + 3(ab + bc + ac) \\ &\ge 3 + 3(3k^2 - 1) = 9k^2 \end{aligned}$$
it follows that $a+b+c \ge 3k$ so $a^3 + b^3 + c^3 - 3abc \ge 9k$.