Find the smallest positive value taken by $a^3+b^3+c^3-3abc$ for positive integers $a,b,c$. Find all integers $a,b,c$ which give the smallest value.
Since it is generally hard to find the minimum of a multivariate polynomial, I tried factoring it at first. The $abc$ reminds me of AM-GM, but I wasn't sure how to relate the $a^3+b^3+c^3$ to it. Also the condition that we are working with integers may make it easier to find the minimum.
Best Answer
$$\quad F=a^3+b^3+c^3-3 a b c=(a+b+c)(a^2+b^2+c^2- a b -b c- c a)=$$ $$= \frac {1}{2}(a+b+c)(\;[a-b]^2+[b-c]^2+[c-a]^2\;).$$ If $a=b=c$ then $F=0.$
If $a,b,c$ are not all equal then $a+b+c\geq 1+1+2=4,$ and also at least two of $|a-b|, |b-c|, |c-a|$ are non-zero, giving $(a-b)^2+(b-c)^2+(c-a)^2\geq 1^2+1^2+0^2=2 .$
Therefore $F>0\implies F\geq \frac {1}{2}(2)(4)=4.$ Which is attained when $a=b=1$ and $c=2.$