I am struggling with an exercise. I am asked to prove the following:
Let $G$ be a simple and undirected graph with a fixed planar embedding such that the circuit bounding the outer face has length at least $\frac{n-5}{2}$. Furthermore, assume that $\chi(G - v) \leq 4$ for all vertices $v \in V(G)$. Then $\chi(G) \leq 4$.
Here is my attempt so far:
First, w.l.o.g. assume that $G$ is $2$-connected: Otherwise pick a vertex $w$ such that $G - w$ is disconnected. Color both connected components with $4$ colors and combine them to a coloring of $G$.
Now, if we manage to prove that there is a vertex $v \in V(G)$ with $\deg(v) \leq 3$ then we are done. Hence, suppose to the contrary that $\deg(v) \geq 4$ for all $v \in V(G)$. Therefore we have $$ m \geq 2n.$$ Since $G$ is 2-connected, every face of the embedding is bounded by a circuit and each circuit has length at least $3$ ($G$ is simple). Hence $$ 2m \geq \frac{n-5}{2} + (f-1)\cdot 3 .$$ My hope is that both inequalities together contradict Euler's formula, but I don't see how. Am I missing something? Or maybe my approach is not promising in the first place? Any help is appreciated.
Edit: Thanks @Math for pointing out an error I made. This is the corrected version following their ideas.
Note that it is sufficient to prove that there exists some vertex $v \in V(G)$ with $\deg(v) \leq 4$ by the proof of the five-color-theorem and the fact that $\chi(G - v) \leq 4$. We assume to the contrary that $\deg(v) \geq 5$ for all $v \in V(G)$ holds. This leaves us with $$ m \geq \frac{5}{2}n.$$
In fact, Euler's formula will yield the result: By Euler's Formula we have $$ 3f = 3m -3n +6. $$ Now, we can plug in the inequalities stated above:
$$ 3f \geq \frac{5n}{2} + \frac{n-5}{2} +(f-1)3 -3n +6 = 3f +1/2$$
and therefore $0 \geq 1/2$, a contradiction.