Problem 2 of Chapter 9 A Concise Introduction to Pure Mathematics

Prove that for a convex polyhedron with V vertices, E edges and F faces, the following inequalities are true:

2E >= 3F, and 2E >= 3V

Proof:

An edge can only be shared by two faces: shared by more than two faces, the polyhedron will not be convex; shared by only one face, there will be a open face, and thus not a solid, not a polyhedron.

A face has at least 3 edges. For each face, e >= 3.

If we take out the faces from the polyhedron one by one, and count their edge. Then total number of edges >= 3F. Total number for edge is 2E in this case, because each edge is share by two faces. Therefore 2E >= 3F.

Similarly, each vertex has at least 3 edges joined together. If we take out the edges one by one, each edge will have 2 vertices, in this case we will count 2E vertices. But because each vertex has at least 3 edges joined together, each vertex is counted at least 3 time. Therefore 2E >= 3V.