I found the following question in a previously assigned graph theory final:
Let $G$ be a simple, planar graph with $1\leq |E(G)|< 30$. Show that $G$ has a vertex with degree at most 4.
My Attempt: this is equivalent to showing $\delta(G)\leq4$. Assume by contradiction that $\delta(G)\geq 5$. First, we know
$5|V(G)|\leq \sum_{v\in V(G)} \text{deg}(v) \leq 2|E(G)|.$
Also, since $G$ is planar we must have
$|E(G)|\leq 3|V(G)|-6,$
which implies
$\frac{5}{2}|V(G)|\leq|E(G)|\leq 3|V(G)|-6.$
From here I am lost. Are there any suggestions for how to continue, or is this not leading to a solution?