I'm working in the following graph theory excercise:
Let $G$ be a connected planar graph of order $n ≥ 5$ and size $m$. Prove that if the length of a smallest cycle in $G$ is $5$, then $m ≤ 5/3(n-2)$.
I'm starting from the theorem that says that while the graph is planar then $m ≤ 3(n-2)$ but I truly don't know how a cycle would be involved in the process, thanks in advance for any hint or help.
Because $G$ is connected and planar, Euler's theorem is bound to be involved. Let $f$ denote the number of faces, so that $n-m+f=2$. Because the length of the smallest cycle in $G$ is $5$, every face has at least $5$ edges adjacent to it. This means $2m\geq 5f$ because every edge is adjacent to two faces. Plugging this in yields $$2=n-m+f\leq n-m+\tfrac25m=n-\tfrac35m,$$ and hence $m\leq\tfrac53(n-2)$.