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. 

From $\displaystyle a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\bigg[(a-b)^2+(b-c)^2+(c-a)^2\bigg]\geq 3.$
Because $a,b,c$ are distinct integers.
So here we have taken $a,b,c$ as $3$ consecutive integers.
And $\displaystyle (a+b+c)^2-3(ab+bc+ca)\geq 0\Rightarrow (a+b+c)^2\geq 3(ab+bc+ca)=3(3k^2-1)$
So $\displaystyle (a+b+c)\geq \sqrt{9k^2-3}$
So we have taken $(a+b+c)\geq 3k.$
So we have $\displaystyle a^3+b^3+c^3-3abc\geq 3\cdot (3k)=9k.$
Equality hold when $a,b,c$ are $3$ consicutive posotive  integers
Namely we have taken  $k-1,k,k+1$
