Degree of vertices in planar graph 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.
 A: 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 
$$k > 6$$
A: First we show that if we have a counter-example, we can find a counter-example which is connected.
Let $G$ be a planar graph with connected components $G_1$, $G_2$, ..., $G_k$.  Select nodes $x_i\in G_i$, and create a new graph $G'$ defined by starting with $G$ and then adding edges $\{x_1,x_2\},\{x_2,x_3\},...,\{x_{k-1},x_k\}$.  I'll leave it to you to prove that $G'$ is still planar, and, if $G$ is a counter-example to your theorem, then so is $G'$.
So, you've shown there can be no connected counter-example with $|G|=12$, and hence, by this argument, there can be no counter-example with $|G|=12$.
[Effort to prove via induction elided.]
As Graphth noted above, your argument can be used for $n=|G|\geq 12$, too.
$$m\leq 3n-6$$
$$\sum_{x\in G} deg(x) = 2m \leq 6n-12$$
But if $G$ is a connected counter-example, then $\sum_{x\in G} deg(x) \geq 8(n-5) + 5$
So $$8n-35 \leq 6n-12$$
or
$$2n\leq 23$$
Note that connectedness is more than you really need, you just need to know that $deg(x)\geq 1$ for all nodes $x\in G$.  So it's very easy to take any counter-example to a counter-example where no nodes are isolated because adding an edge from an isolated node to another node does not break the planar nature of the graph.
