Given a 3D convex polyhedron $P$ with $n$ vertices, when $n$ is large, is there a tight upper bound (or at least one that holds "most of the time") on the number of (triangular faces) $P$ has?
Here triangular faces means if a face isn't triangle then we break it into triangles. For example a face that's a square or rectangle breaks down into two triangular faces.
Of course $C_n^3$ is a bound but it's too loose. It'd be nice if there's a bound of lower order.
A convex polyhedron with triangular faces and $n$ vertices has exactly $2n-4$ faces.
This follows from Euler's formula $V-A+F=2$ because $2A=3F$. Indeed, every face has three edges which are counted twice since each edge belongs to exactly two faces.