If $ab+bc+ca \geq 3k^2-1$, prove that: $a^3+b^3+c^3-3abc \geq 9k$. 
If $ab+bc+ca \geq 3k^2-1$, prove that: $a^3+b^3+c^3-3abc \geq 9k$.

I recently came across a question in which we had to prove the above inequality using the given condition as mentioned above. Here $a,b,c$ are distinct positive integers and $k$ is also a positive integer. I absolutely have got no idea how to solve it or efficiently use the condition 'positive integers'. Furthermore, although the expression $a^3+b^3+c^3-3abc$ seems a bit familiar but I'm not able to understand how to make the condition useful. 
Please help. 

Use that $$a^3+b^3+c^3-3abc=\left( a+b+c \right)  \left( {a}^{2}-ab-ac+{b}^{2}-bc+{c}^{2}
 \right) 
$$
