Minimum Degree Bound for Simple Planar Graph

Theorem
Let $G$ be a simple connected planar graph. Then, $\delta \left(G \right) \leq 5$, where $\delta$ is a minimum degree of a graph.

Linear Bound on Edges
First, we will show that the linear bound on edges for any simple connected planar graph $G$ with $n \geq 3$ vertices is exactly:
 * $m \leq 3n − 6$, where $m$ is a number of edges.

Let sequence $s_1,. . ., s_f$ be the regions of a planar embedding of $G$. We denote the number of boundary edges of $s_i$ by $r_i \:\: \left(i = 1, . . ., f \right)$.

In case $f = 1$, $G$ is then a tree and $m = n − 1 \leq 3n − 6$.

Thus, we assume that $f \geq 2$. Since $G$ is simple, then (by the definition of planar embedding): Using this two facts, we can find the boundary for $\sum\limits_{i+1}^f r_i$ as
 * every region has at least $3$ boundary edges,
 * and every edge is a boundary edge of one or two regions in the planar embedding.
 * $3f \leq \sum\limits_{i+1}^f r_i \leq 2m$

Now calculating the Euler Polyhedron Formula with $f \leq 2m/3$, we will arrive to $m \leq 3n − 6$.

Degree Bound
Now when we have linear bound on edges, we can prove the main theorem by contradiction. Consider the counter hypothesis:
 * $G$ is a simple planar graph and $\delta \left(G \right) \geq 6$.

Then, by the Handshake Lemma $m \geq 3n$, where $n$ is the number of vertices and $m$ is the number of edges in $G$. This contradicts the linear bound on edges.