Question

what are the minimum number of faces a polyhedron can have?

Answer 1

It have more the 8 ok

Answer 2

The easiest way to prove that n+1n+1faces are necessary is probably to show that given only nn vectors — representing the normals to the nn faces of the polytope — there's always a non-zero nn-vector that has either zero or positive dot product with all of them; such a vector will therefore 'go off to infinity' staying within the half-plane defined by all of the faces, and thus witnesses the unboundedness of the polytope. This can be done via a relatively straightforward application of Gaussian Elimination.