This is a problem I have had for a while. For a triangle, the side opposite the largest angle has the largest length (and similarly for smallest angle). For a tetrahedron, the question is whether the face opposite the largest solid angle (trihedral angle - the area of the spherical triangle with angles equal to the dihedral angles incident on one vertex) has the largest area?
I can prove that this is true for tetrahedra coming from sphere packings (each vertex $v_i$ has a weight $r_i$ and the edge lengths are equal to $r_i+r_j$, etc.) but this excludes degenerations of a tetrahedron such as the one going to a box with two diagonals.
I suspect this is true but have no proof. Perhaps someone else does?