If degenerate triangles were allowed, the optimal choice would be a collinear triangle. For a non-degenerate triangle we want to come as close as possible to collinearity; triangles with sidelengths 25, 26, 49 or 2, 49, 49 come to mind. Let us try to produce a rigorous argument for the claim that the latter sidelengths 2, 49, 49 do indeed produce the smallest possible area.

We compare the areas of a triangle with sidelengths and a triangle with sidelengths , assuming that and . Using Heron's formula, we see that it suffices to compare the terms
for the first triangle and
for the second triangle. As the first two factors are the same in both terms, we can reduce the comparison to the terms
and
Subtracting the last term from the one before it, we get , which is positive due to . This shows that the area of the second triangle is smaller.

Now, if we have a non-degenerate triangle with sidelengths and and and with minimal area, the above step from sidelengths to sidelengths must be barred for some reason (otherwise the area would not be minimal). The only possible reason for that is that a non-degenerate triangle with sidelengths does not exist due to a violation of the triangle inequalities. As the triangle inequalities for the ---triangle are fulfilled, the only triangle inequality which can be violated for the hypothetical triangle with the new sidelengths is , meaning that in fact we have
As the triangle inequalities for the ---triangle are fulfilled (due to its existence), we have and therefore
So, taking these inequalities together, we have
and, adding 1, we get
Consequently, is equal either to or to . Assuming the first case, , we get
which is not possible for an integer value of . So the second case must be true: . This leads us to
which means
and We are almost done. We still need to find out what and are. As we already know that , the Heronian formula tells us that the area of our minimal triangle only depends on the term
which is equivalent to As is already fixed (), the only variable here is the product . Now
For the positive integer value of , we need to get as far away from as possible to minimalize . As , we must investigate the lower end of the spectrum for . The value is forbidden, as would be then, violating and the triangle inequality . The best acceptable choice is obviously . So we finally have