TITLE: Degree of vertices in planar graph
QUESTION [7 upvotes]: Here is the problem:

Let $G$ a planar graph with $12$ vertices. Prove that there exist at
  least $6$ vertices with degree $\leq 7$.

Here it is what I did:
Since $G$ is planar the number of its edges is $ m\leq 3n-6=30$.
Assume now that there are only $5$ vertices ($w_1 ,w_2 ,\cdots ,w_5$) with degree $\leq 7$. Then the other $7$ vertices have degree $ \geq 8$.
It is  $2m= \displaystyle{ \sum_{v \in V(G)} deg(v) \geq \sum_{j=1}^{5} deg(w_j) +7 \cdot 8}$
so it must be  $\displaystyle{\sum_{j=1}^{5} deg(w_j) +56 \leq 60}$.
From here I can't do nothing.

REPLY [6 votes]: Let $G$ be a planar graph with $n$ vertices and the minimum number $k$ vertices of degree at most $7$.  We may assume that $G$ is maximal, since adding edges doesn't increase the number of "small degree" vertices.
In this case, if $n>3$ there are no vertices of degree two, since a path going through a degree two vertex can't be in two faces bounded by three edges.
From Euler's formula:
$$6n - 12 \ge 3k + 8(n - k)$$
or 
$$5k \ge 2n + 12$$
If $n=12$ we discover
$$5k\ge 36$$
and so 
$$k > 6$$