I am trying to develop an integral in $R^n$ and I am faced with the following problem:
Given a polyhedron $P$ in $R^n$ of diameter d, define the "compactness" of $P$ as the quotient of the volume of $P$ by $d^n$. It is clear that that the compactness is invariant under a change of scale.
I am certain that the following assertion is true, but it is difficult for me to prove it:
"Among all the 2-d polyhedra with 4 vertices, the square owns the largest compactness; among all the 3-d polyhedra with 8 vertices, the cube owns the largest compactness, and in general, the hypercube is the unique n-d polyhedron of $2^n$ vertices whose compactness is maximal."
Theoretically, it should be possible to solve this problem using differential calculus, but such a direct approach seems awful. A more clever way would be to seek the maximal polyhedron of $2^n$ vertices whose diameter is given, using, say, Lagrange multipliers, but this seems awful too.
My impression is that a proof should begin with: "Assume that the polyhedron is not a cube. Then by moving a bit some vertex(es), it is possible to obtain a polyhedron with the same diameter and with larger volume".
N.B: 1.the volume of the n-d polyhedron is the sum of the volumes of the n-d simplexes composing it. 2. the diameter of a polyhedron is necessarily the largest distance between two vertexes, among all the pairs of vertexes.