Let $G$ be a planar graph with no triangles and $m\geq 2$ edges and $f$ faces. Show that $G$ contains a vertex with degree $\leq 3$.
In a previous exercise, I showed that $f\leq m/2$. I tried proving this by induction on the number of edges. Say we remove an edge $\{u,v\}$. Then if either $u$ or $v$ has degree $<3$, we know that $G$ has a vertex with degree $3$. But I’m not really sure how to proceed, and what to do with the fact that there are no triangles.
Suppose $d_i\geq 4$ for each $i$. Then by handshake lemma we have $2m\geq 4n$ where $n$ is a number of vertices.
Now using the fact (you proved) $f\leq m/2$ and Euler formula for planar graphs
$$ 2+m =n+f \leq m/2+m/2 =m$$ we get a contradiction.