Sum of dihedral angles in Tetrahedron I'd like to ask if someone can help me out with this problem. I have to determine what is the lower and upper bound for sum (the largest and smallest sum I can get) of dihedral angles in arbitrary Tetrahedron and prove that. I'm ok with hint for proof, but I'd be grateful for lower and upper bound and reason for that.
Thanks
 A: Lemma: Sum of the 4 internal solid angles of a tetrahedron is bounded above by $2\pi$.
Start with a non-degenerate tetrahedron $\langle p_1p_2p_3p_4 \rangle$. Let $p = p_i$ be one its vertices and $\vec{n} \in S^2$ be any unit vector. Aside from a set of measure zero in choosing $\vec{n}$, the projection
of $p_j, j = 1\ldots4$ onto a plane orthogonal to $\vec{n}$ are in general positions (i.e. no 3 points are collinear). When the images of the vertices are in general positions, a necessary condition for either $\vec{n}$ or $-\vec{n}$ belong to the inner solid angle at $p$ is $p$'s image lies in the interior of the triangle formed by the images of other 3 vertices. So aside from a set of exception of measure zero, the unit vectors in the 4 inner solid angles are "disjoint". When one view tetrahedron $\langle p_1p_2p_3p_4 \rangle$ as the convex hull of its vertices, the vertices are extremal points. This in turn implies for any unit vector, $\vec{n}$ and $-\vec{n}$ cannot belong to the inner solid angle of $p$ at the same time.
From this we can conclude (up to a set of exception of measure zero), at most half of the unit vectors belongs to the 4 inner solid angles of a tetrahedron. The almost disjointness of the inner solid angles then forces their sum to be at most $2\pi$.
Back to original problem
Let $\Omega_p$ be the internal solid angle and $\phi_{p,i}, i = 1\ldots 3$ be the three dihedral angles at vertex $p$. The wiki 
page mentioned by @joriki tell us:
$$\Omega_p = \sum_{i=1}^3 \phi_{p,i} - \pi$$
Notice each $\Omega_p \ge 0$ and we have shown $\sum_{p}\Omega_{p} \le 2\pi$. We get:
$$\begin{align}
         & 0 \le \sum_p \sum_{i=1}^3 \phi_{p,i} - 4\pi \le 2\pi\\
\implies & 2\pi \le \frac12 \sum_p \sum_{i=1}^3 \phi_{p,i} \le 3\pi
\end{align}$$
When we sum the dihedral angles over $p$ and $i$, every dihedral angles with be counted twice. This means the expression $\frac12 \sum_p \sum_{i=1}^3 \phi_{p,i}$ above is nothing
but the sum of the 6 dihedral angles of a tetrahedron.
