Let $\Omega$ be an $n$-simplex, which means the convex hull of $(n+1)$ points in $\mathbb{R}^n$. We denote $P(\Omega)$ be the perimeter of $\Omega$ and $|\Omega|$ be the volume of $\Omega$.
For example, a $2$-simplex is a triangle in $\mathbb{R}^2$ and a $3$-simplex is a tetrahedron in $\mathbb{R}^3$. The perimeter of a $2$-simplex is the usual perimeter of a triangle, while the perimeter of a $3$-simplex is the surface area of the tetrahedron.
Now my question is, if we want to minimize the perimeter over all $n$-simplexes with the same volume, does the regular $n$-simplex has least perimeter?
I can show that if $n=2$, this is true. Indeed, presrcibing area of a triangle, the equilateral triangle has least perimeter, which is standard.
If $n=3$, the same result might also be true, since I found a reference here. However, I think there is a gap in the proof.
My question is whether it is always true that the regular $n$-simplex is the unique shape to minimize perimeter, among $n$-simplexes with the volume. Is this a classical result?
Any help would be really appreciated.
I have found the correct reference:
Mitrinovic, D. S., Pecaric, J. E. and Volenec, V.: Recent Advances in Geometric Inequalities, Kluwer, Dordrecht, Boston, London, 1989.
The desired result is in Page 519, after applying arithmetic-geometric mean inequality.