Show that the graph K_{3,3} is not planar.

My working:

Assume that K_{3,3} is planar.

So now every cycle has at least 4 edges, so every face is at least bounded by 4 edges, thus the number of edges bounding faces is 4f.

In planar graphs, each edge belongs to at most 2 bounding cycles.

Now I am stuck, my textbook says that we can form this inequality 2e \ge 4f, I'm sure this is something very obvious but I can't see how the inequality is formed. The RHS of the inequality counts the least number of edges which bounds the total number of faces and the LHS counts the total number of bounding cycles which each edge belongs to. How is the LHS and RHS related? How is the inequality then formed?
Thanks!