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.