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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.

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Yes thats the correct answer. We can do it as $3(ab+bc+ca)>321$ and for the value to be non negative (a+b+c)^2 should be just greater than 321 so we can suppose it to be 18 as 18^2=324 and hence (a+b+c)>18 for minimum value it should 18. and also abc should be greatest for minimum value so we struck to values 5,6,7 . and the minimun value will be 54

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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.

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