Minimum value of the expression given below. Is $a,b,c$ are three different positive integers such that :
$$
ab+bc+ca\geq 107
$$
Then what is the minimum value of $a^3+b^3+c^3-3abc$.
I expanded this expression to $(a+b+c)((a+b+c)^2-3(ab+bc+ca))$ and tried to find the minimum value of $(a+b+c)$ by AM-GM inequalities but the problem is that the minimum value occurs at scenario where $a=b=c$, which is ommitted by assumptions of this question.
I'd appreciate some help.

You can see $a^3+b^3+c^3-3abc=\frac{1}{2}[(a-b)^2+(a-c)^2+(c-b)^2](a+b+c)$
For $[(a-b)^2+(a-c)^2+(c-b)^2]$, since they are three different integers, this is greater or equal to $[(0-1)^2+(0-2)^2+(1-2)^2]=6$;
For $(a+b+c)$, $(a+b+c)=\sqrt{(a+b+c)^2}\geq \sqrt{3(ab+bc+ca)}\geq\sqrt{3*107}>17$. Since the're integers, we have $a+b+c\geq 18$.
Check for $a=5, b=6, c=7$, should be the minimum point.
