The $3$–$4$–$5$ triangle has area $\frac{3 \cdot 4}{2} = 6$. We will prove that no other triangle with integer sidelengths and area has smaller area.
Let $a$, $b$, $c$ be the sidelengths. Then its area is $S = \sqrt{s(s-a)(s-b)(s-c)}$, where $s = \frac{a+b+c}{2}$. Since the area is also an integer, $a+b+c$ is even, and $s$, $s-a$, $s-b$, $s-c$ are all integers.
Now, notice that the triangle cannot be equilateral, since equilateral triangles with an integer side have irrational area. So, at least two of the three integer numbers $s-a$, $s-b$, $s-c$ are distinct and, since $s = (s-a) + (s-b) + (s-c) \ge 1+1+2 = 4$, $S \ge \sqrt{4 \cdot 2 \cdot 2 \cdot 1} = \sqrt{8}$, so $S \ge 3$.
If $S$ is odd, $s \ge 5$ and $s-a$, $s-b$, $s-c$ are all odd, so $S \ge \sqrt{5 \cdot 3 \cdot 1 \cdot 1} = \sqrt{15}$, so $S \ge 5$. The only relevant case is $S = 5$. But this would imply two of $s$, $s-a$, $s-b$, $s-c$ being equal to $5$, which is impossible.
If $S$ is even, the only relevant case is $S = 4$. But then all of $s$, $s-a$, $s-b$, $s-c$ are powers of two. So if $s > 4$ then $s \ge 8$ and $s-a$, $s-b$, $s-c$ would be, in some order, $1$, $1$, $2$, which is not possible because $s = (s-a)+(s-b)+(s-c)$. If $s = 4$, the only possibility would be $s-a$, $s-b$, $s-c$ being $1$, $1$, $2$, which does not work either.