Bipartite Planar Graph

Question

Prove that a bipartite (ie, two-colorable) planar graph with $E$ edges and $V$ vertices satisfies: $E\leq 2V-4$, for $V\geq 3$.

Answer 1

Consider an embedding of $G$. Note that since $G$ is bipartite then it is triangle free. Note that the walk around any face should have length at least 3 (assuming the graph is simple). Now, note that if the walk around a face has length 3, then it must be a triangle. Therefore the walk around any face has length at least 4. Now, let $\mathcal{F}$ denote the set of faces in the given embedding and let $l(f)$ denote the length of the walk around face $f\in\mathcal{F}$. Observe that $2E=\sum_{f\in\mathcal{F}}l(f)\geq 4F$. Therefore $E\geq 2F$. Now, by Euler's formula $2=V-E+F\leq V-E+\frac{E}{2}$. This can be rearranged to the desired inequality.

Update: Was assuming each face was bounded by a cycle. New argument works even after dropping this assumption.