Connected planar simple Graph: number of edges a function of the number of vertices Suppose that a connected planar simple graph with $e$ edges and $v$ vertices contains no simple circuit with length greater than or equal to $4.\;$ Show that $$\frac 53 v -\frac{10}{3} \geq e$$
or, equivalently, $$5(v-2) \geq 3e$$

Try using Euler's "polyhedral formula" - If G is a connected plane graph then V + F - E = 2.
