Simple connected planar graphs and the relationship between vertex degree and number of edges.

Question

I'm currently studying on planar graphs and in the textbook there's the following problem. *Given a simple connected planar graph G, we have to prove that if graph G has less than 30 edges then it must have at least one vertex of degree at most 4.* I understand I have to use Euler's formula where $f=m-n+2$ together with the formula for max # of edges $m\leq3n-6$. However, I'm getting confused. I imagine I have to end up in a inequality something like $30\leq3n-6$ and combine this with Euler's formula. Can anyone give me a hand on this?

Answer 1

Consider a planar graph with $29$ edges; the $m\le3n-6$ inequality means $n\ge12$. But then the average degree is at most $\frac{58}{12}<5$, which means there must be at least one vertex with degree $4$ or less.

The bound of $30$ edges is tight: the skeleton of the regular icosahedron has that many edges, is planar and has no degree-$4-$ vertex.