Is the average number of edges per 2-faces of a convex 3-polytope always below six?

Question

Is the average number of edges per 2-faces of a convex 3-polytope always below six? Which theorem can answer this question or how do you prove this?

Answer 1

Old man Euler will do. $$V-E+F=2$$

Now suppose the average number of edges per face is some $r\geqslant6$. Then count the edges of your polyhedron, iterating through all faces. In doing so, you will count each edge twice (for it belongs to 2 faces), hence don't forget to divide by 2. All in all, you will end up with $$E={r\over2}\cdot F$$

OK, now move on to vertices. Each face contains as many vertices as edges, but each vertex belongs to at least 3 faces, or maybe even more. Hence $$V\leqslant{r\over3}\cdot F$$

Now put it all together.

$$2 = V-E+F \leqslant\left({r\over3} - {r\over2} + 1\right)\cdot F = {6-r\over r}\cdot F\leqslant0$$