Can you find a tetrahedron T contained inside a tetrahedron W such that the sum of the lengths of the edges of T is bigger than the sum of the lengths of the edges of W?
Let ABCD be tetrahedron T. Make AB=BC=CA=1 and position D slightly above the center of triangle ABC. The sum of edge lengths of T is then slightly larger than 3+sqrt(3)=4.732.
Let point E be very close to A on edge AD. Then let tetrahedron W be ABCE. The sum of edge lengths of T is then very close to 5.
Since 5 > 4.732 the sum of edge lengths of tetrahedron W is greater than that of T but W is contained within T.
Solution
