Let $K\subset \mathbb R^3$ be a non-degenerate tetrahedron. This tetrahedron has the property that the ratio between diameter and radius of insphere is bounded, $$ d \le \kappa r\ \text{ for some } \ \kappa>0 $$ where $d$ is the diameter of $K$, whereas $r$ is the radius of the insphere.
Is it possible to derive a lower bound on the solid angels of the tetrahedron in terms of $\kappa$ only?
The purpose of such an estimate would be as follows: Let tetrahedrons $K_i$ be given, each satisfying the inequality above. Assume that they do not overlap and share one common vertex. Then the number of such tetrahedrons is bounded from above by a finite number.
I could easily prove the analogous result for triangles. The desired bound is Zlamal's minimum angle condition in this case.
Such an assumption as above is traditionally assumed in finite element methods. I could not find any proof of the above statement (bounding the number of tetrahedrons with one common vertex), which is in most references mentioned as fact.
Let $x_0$ be the vertex of the angle in question, $x_i$ the midpoint of the insphere, and set $L:=|x_i-x_0|$. Then take a ball centered at $x_0$ with radius $\sqrt{L^2+r^2}$. The surface area $S$ of the projection of the tetrahedron onto this ball is $(r^2+L^2)\cdot \Theta$ with $\Theta$ being the solid angle. This area is larger than the area $S_i$ of the projection of the insphere onto the ball. The area $S_i$ is the surface of a spherical cap whose base circle is the circle with radius $r$ and center $x_i$. Hence, $S_i$ is larger than the area of this circle. Since the insphere lies inside the tetrahedron, it holds $\sqrt{L^2+r^2}\le d$, which implies $$ \pi r^2 \le S_i \le S = (L^2+r^2)\cdot \Theta \le d^2 \cdot \Theta. $$ Now the left-hand side can be bounded from below by $\pi \kappa^{-2}d^2$, which leads to the estimate $$ \Theta\ge \pi \kappa^{-2}. $$