Proof that every finite planar graph contains at least one vertex with valence less than six

Assume, to the contrary, that

there is a finite planar graph with all vertices having valence >= 6.      (1)

Let V,E,F be the number of vertices, edges, faces respectively.

Then 2*E = SUM(vertices v) valence of v

         >= SUM(vertices v) 6           since each valence >= 6 by (1)

         = 6*V

I.e.   V <= E/3                                                            (2)

Also, 2*E = SUM(faces f) valence of f

         >= SUM(faces f) 3      (equality if the graph is triangulated)

          = 3*F

I.e.    F <= 2*E/3                                                         (3)

Euler's formula says:

        2 = V - E + F

         <= E/3 - E + 2*E/3             by (2) and (3)

          = 0

Contradiction.