Optimaximization

Let's consider a case when three of "our edges whose lengths sum to 1" belong to one face. To maximize the volume we need maximum area for a given perimeter and maximum height of the tetrahedron. That will be achieved with the base being an equilateral triangle because it has a maximum ratio of area to perimeter, and the fourth edge with the required restriction to be perpendicular to the base. Let $x$ denote the length of the base edge, then the volume will be: $V=\frac{\sqrt{3}}{4} x^2 \times (1-3x) \times \frac{1}{3}$ After differentiating the above function we find the local maximum to be at $x=2/9$ which gives volume $V=\frac{1}{3^5\sqrt{3}}\ \Rightarrow A+B=\boxed{8}$

Comment: As Mark Hennings correctly points out in the comment below, there is another configuration when no three edges with the given restriction belong on one face.