Can a tetrahedron lying completely inside another tetrahedron have a larger sum of edge lengths? Find 2 tetrahedrons $ABCD$ and $EFGH$ such that


*

*$EFGH$ lies completely inside $ABCD$.

*The sum of edge lengths of $EFGH$ is strictly greater than the sum of edge lengths of $ABCD$.
I am completely stumped on this. Seems very counter intuitive to begin. I now have doubts if a solution exists or not.
Source : Here
 A: $$A=(0,0,0), \;B = (1,0,0),\;C=(1,1,0) ,\;D=(-1,0,1)$$
$$E =(1,1-\epsilon,0),\; F =  (1-\epsilon,0,\epsilon/2),\; G = (-1+\epsilon,0,1-\epsilon/2) ,\;H = (-1+\epsilon,\epsilon/2,1-\epsilon/2)$$
Let $\epsilon = 0.01$, sum of edge length of $ABCD$:
$$
|AB|+|AC|+|AD|+|BC|+|BD|+|CD| = 2+2\sqrt{2}+\sqrt{5}+\sqrt{6}\approx 9.51.
$$
sum of edge length of $EFGH$:
$$
|EF|+|EG|+|EH|+|FG|+|FH|+|GH| \approx 10.3.
$$
Roughly looks like the following:

You can perturb $EFGH$ by setting them completely inside $ABCD$, by adding another parameter $\delta\ll 1$ while having the sum of edge lengths changing by $O(\delta)$. 
The backstory of this construction is that: If $ABCD$ has three long edges, we can make $EFGH$ having four long edges.
A: Think of a tetrahedron composed of $(-k,0,0), (-k,\epsilon,0), (0,\epsilon,0), (0,0,k)$ where $k$ is large and $\epsilon \ll 1$  You can move $(0,\epsilon,0)$ to the right (and down a bit) and increase the sum of the sides.  You can then move all the points inward by $\epsilon^2$ to get strictly inside.
A: A spire is a tetrahedron with one tiny face and three long edges. A splinter is a tetrahedron with two tiny opposite edges and four long edges.

From Cheng et al., "Sliver Exudation", Proc. J. ACM, 2000.
Take a spire of height $1$ and fit a splinter inside it. The sum of edge lengths of the spire is $\approx 3$ while that of the splinter is $\approx 4$.
