Suppose that G is a connected simple planar graph on at least three vertices. Let F denote the number of faces of G, and E the number of edges. Prove that 3F ≤ 2E.
Using (1), prove that if G is a connected simple planar graph on n ≥ 3 vertices, then G contains at most 3n−6 edges.
Show that the complete graph on five vertices, $K_5$, is not a planar graph.
Can anyone help me with above questions? Really cant seems to find a way to solve it.
Every face is made of at least three edges, and every edge is a part of two faces. From that, you need to make an inequality
You need to connect the previous inequality with well known $n-m+F=2$ equality, which can be proved by induction
This goes from previous point