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.