Find the smallest positive value taken by $a^3+b^3+c^3-3abc$

Question

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.

Answer 1

The answer is $4$, for $a = 1, b = 1, c = 2$ (or some permutation).

We have $$a^3 + b^3 + c^3 - 3abc = (a + b + c)[(a + b + c)^2 - 3(ab + bc + ca)].$$ This shows that any positive value taken by the expression must be at least as large as $a + b + c$, which leaves only the above cases and $a =b = c = 1$ to examine.

Edit Let $A = a^3 - 3abc + b^3 + c^3$. We will write $A$ in terms of the elementary symmetric polynomials. The leading term in lexicographic order is $a^3$, so we compute $$A - (a + b + c)^3 = - 3a^2b - 3a^2 c - 3ab^2 - 9abc - 3ac^2 - 3b^2 c.$$ The leading term in this expression is $-3a^2b$, so we compute $$A - (a + b + c)^3 + 3(a + b + c)(ab + ac + bc) = 0.$$

Answer 2

$$\quad F=a^3+b^3+c^3-3 a b c=(a+b+c)(a^2+b^2+c^2- a b -b c- c a)=$$ $$= \frac {1}{2}(a+b+c)(\;[a-b]^2+[b-c]^2+[c-a]^2\;).$$ 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 $c=2.$