Planar graph with V ≥ 2 has at least 2 vertices whose degrees are at most 5

Question

If G was a planar graph on V ≥ 2 vertices. How would I go about proving that G has at least 2 vertices whose degrees are at most 5?

I understand that planar graphs can be drawn so that every edge is represented by a continuous curve and no two edges intersect except at the endpoints.
I know that if |V| ≥ 3 vertices then 3|V| - 6 ≥ |E|. But I am not sure how to use that information ? Could someone guide me through this process please? and thanks for the help

Answer 1

I assume that in your terminology "graph" precludes loops and parallel edges, otherwise the claim is false.

Let $v_k$ be the number of vertices of degree $k$. Then the total number of vertices and the total number of edges are given by $$v=v_0+v_1+v_2+v_3+\cdots\quad\hbox{and}\quad 2e=v_1+2v_2+3v_3+\cdots\ .$$ Substituting into $6v\ge12+2e$ and rearranging gives $$6v_0+5v_1+4v_2+3v_3+2v_4+v_5\ge12+\cdots+v_7+2v_8+3v_9+\cdots\ge12\ .$$ But the only way the LHS can be $12$ or more is if $v_0\ge2$, or at least three of the numbers $v_0,\ldots,v_5$ are at least $1$.