Suppose G is a connected planar simple graph with $e$ edges and $v$ vertices with no cycles of length 4 or less.
Prove that (For $ v ≥ 4$): $$e ≤ {\frac{5}{3}}v - {\frac{10}{3}}$$
-I have a basic understanding of planar graphs based off of our lecture in class. I understand that planar means it can be drawn on a single plane without any edges crossing and I know that since it's a connected graph it has one component, and I know that since it is a simple graph it does not contain loops or multi-edges. Although I don't understand how to tie this in with a proof such as this, any help is appreciated.
The following theorem is well known in graph theory (see theorem 6.10 in http://compalg.inf.elte.hu/~tony/Oktatas/TDK/FINAL/Chap%206.PDF):
For a connected simple planar graph with $v$ vertices and $e$ edges, and girth $g$, we have $$e\leq \frac{g}{g-2}(v-2)$$
Note the girth $g$ is the length of the smallest cycle.
In our case $g=5$, thus $$e\leq \frac{5}{3}(v-2)$$