Austrian Mathematical Olympiad

It is well-known and easily verified that $$a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)((a-b{)}^2+(b-c{)}^2+(c-a{)}^2).(1)$$ 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.