Question: Consider two tetrahedrons one placed inside the other. 
Prove that the sum of all 6 sides of inner tetrahedron is at most the sum of the 6 sides of exterior tetrahedron.

Answer: 
The conjecture about the quotient of the sums of the lengths of edges for a tetrahedron inside another was formulated (a long time ago) by Jerzy Browkin, who saw that, somewhat paradoxically it can be greater than 1, and even arbitrarily close to 4/3, as his examples proved it. He asked if   $\frac 43$   is the least upper bound of such quotients.

The conjecture was solved in 1961 by Wlodzimierz Kuperberg and myself. We published it in 1962 or 1963, in Polish, in Wiadomosci Matematyczne (of Polish Mathematical Society). Years later an English translation was published in Alabama Journal of Mathematics (or similar). And a powerful complete multidimensional generalization was obtained and published by Carl Linderholm, [An inequality for simplices](http://link.springer.com/article/10.1007%2FBF00147532) .

Ancient Greeks could easily formulate and solve the given problem (in 3d) but (as long as we know) never did--it took Jerzy Browkin to raise it.
Sum of the 6 edges of the inner tetrahedron can be larger than the sum of the 6 edges of the outer tetrahedron.
Imagine the outer tetrahedron with 3 sides of 1000 and 3 sides of 1. The inner tetrahedron has 2 vertices close to the top and 2 vertices close to the bottom.
Sum of outer = 1000+1000+1000+1+1+1=3003
Sum of inner = 990+990+990+990+1+1= 3962

[enter image description here](https://i.stack.imgur.com/GTpzV.png)