A tetrahedron inside another tetrahedron. Could the contained tetrahedron have a greater perimeter then the outside one? Suppose you have a tetrahedron. It doesn't have to be regular. Now suppose you have another tetrahedron contained inside the first tetrahedron. Again do not assume it is regular and do not assume that both tetrahedrons are similar. Could it happen that the perimeter of the tetrahedron in the inside is larger then the tetrahedron on the outside? Whether it is or not prove it. Do a proof by contradiction.
This one is tricky. I have been thinking about this problem for a long time now and do not see any way to come up with a reasonable solution. The help would be greatly appreciated! 
 A: The tetrahedron with vertices
$$
(1,0,0)
\qquad (\cos\tfrac{2\pi}3,\sin\tfrac{2\pi}3,0)
\qquad (\cos\tfrac{4\pi}3,\sin\tfrac{4\pi}3,0)
\qquad (0,0,0)
$$
has perimeter $3\sqrt3+3$ and contains the tetrahedron with vertices
$$
(1,0,0)
\qquad (\cos\tfrac{2\pi}3,\sin\tfrac{2\pi}3,0)
\qquad (\cos\tfrac{4\pi}3,\sin\tfrac{4\pi}3,0)
\qquad (1,0,0)
$$
which has perimeter $5\sqrt3$.
(Both tetrahedra are degenerate, for ease of calculation; for a nondegenerate example, replace $(0,0,0)$ in the outer tetrahedron with $(0,0,\varepsilon)$ and replace one of the $(1,0,0)$ in the inner tetrahedron with $(1-\delta)(1,0,0) + \delta(0,0,\varepsilon)$, where $\varepsilon$ and $\delta$ are positive and small.)
Outer tetahedron, roughly:

Inner tetrahedron, roughly:

A: Taking 2-dimensional case, call $ \Sigma $ sum of edge lengths.
Divide each length in ratio p: (1-p) where  p < 1. 
Reduced perimeter of inside triangle $ \Sigma_p $ < $ \Sigma $ 
Find a point inside tetrahedron and use parallel similitude projection to scale q < 1 
on new parallel plane.
$ \Sigma_{p q} $ < $ \Sigma_p $ < $ \Sigma $.
The above is valid for any projection of the inside tetrahedron face triangles subset.
It is true for the superset by simple mathematical induction.
