TITLE: convex polytopes with many faces and edges but few cells and vertices 
QUESTION [6 upvotes]: For a convex polytope $P$ in $\mathbb R^4$, denote by $N_0,N_1,N_2,N_3$ respectively the number of vertices, edges, faces, cells. By Euler's formula, we know $N_0+N_2=N_1+N_3$, which means there is a sort of equilibrum among the $N_i$. But I wonder if there exist upper and/or lower bounds for $f(P):=\frac{N_1+N_2}{N_0+N_3}$ where  $P\subset\mathbb R^4$ is any convex polytope.
E.g. for the regular one called 24-cell, we have $f(P)=4$.

REPLY [3 votes]: Each cell is a convex polytope in 3-space and so has at least 4 faces, whereas each face should have only 2 cells, or $N_3\geq 2N_4$. The same argument for the dual polytope gives $N_1\geq 2N_0$, which gives a lower bound of $f\geq 2$, attained by the $5$-simplex.
This is analogous to the polyhedral bound of $2E \geq 3F,3V$.