# Given a planar graph with minimum cycle length 8, show that $|E| \le \frac{4}{3}|V| - \frac{8}{3}$

Here's what I've got so far. I'm stuck on how to proceed. I believe I need to plug back into Euler's formula, but I'm not getting what I'm looking for by doing that. Where is the denominator of $3$ coming from in the result? Can you please check over my proof for any incorrect statements, and help me move forward?

Proof. Let $G$ be a connected planar graph with at least two edges, and a cycle with length $\ge 8$. Pick a crossing-free embedding of $G$; this embedding has $f$ faces. By Euler's formula, $$f = 2 - |V| + |E|\;.$$ We calculate the sum of the degrees of the faces in this embedding. On the one hand, the sum of the face degrees is $2|E|$ by proposition. On the other hand, since there is a cycle of at least length $8$, the sum of the face degrees is at least $8f$.

That's where I'm stuck at. Any ideas?

If every cycle has length at least eight, then every face in your crossing-free embedding is bounded by at least eight edges, so $8f < 2|E|$, or simply $4f < |E|$. Combining this with Euler's characteristic formula, \begin{align} |E|=f+|V|-2 &\implies 4|E| = 4f + 4|V| - 8 \\ &\implies 4|E| \leq |E| + 4|V| - 8 \\ &\implies 3|E| \leq 4|V| - 8 \\\\ &\implies |E| \leq \frac{4}{3}|V| - \frac{8}{3} \\ \end{align}