Prove that a graph is a maximal planar graph if and only if $e = 3v − 6$ Definition: A planar graph with no multi-edges $G$ is called a maximal planar graph if the graph formed by addition of any edge (not already in the $G$) is not planar or the graph is $K_3 $ or $K_4$ (in these cases we can't add any more edges!)
I am asked to prove that a graph is maximally planar if and only if $e = 3v - 6$.
To prove the first direction (If the graph is maximal then $e = 3v-6$)
$\longrightarrow$
I use the fact that every face in a maximal planar graph is a triangle and that every edge is incident to exactly two faces, and so we have $$ 3f = 2e \to f = \frac{2e}{3}$$
Now I substitue this value of $f$ Euler's formula $$v + f = e + 2$$ to get $$v + \frac{2e}{3} = e  + 2$$ and so we have $$e - \frac{2e}{3} + 2 = v$$ and so we have $$\frac{e}{3} = v - 2$$
and so we finally get $$e = 3v - 6$$
$ \longleftarrow$
Now I have some troubles in proving that other direction, that is if $e = 3v-6$ then the graph if maximal planar. I was thinking by mathematical induction on the number of vertices. But I am lost on how to even begin, what will be the least pair of $(e,v)$ for this to work, I assume $v=1,2$ both doesn't make any sense and so $v=3$ will be the least value, and if we have $v = 3$ then we would have $e= 3 \times 3 - 6 = 3$ so we would have three edges for $3$ vertices which is basically $k_3$ and so it works as a base case. 

Assume $v\geq 3$ and fix an embedding of a planar graph and let $f_i$ be the number of faces with $i$ edges along its boundary. Since every edge is on the boundary of two (not necessarily distinct) faces  we have \begin{align*} 2e&=3f_3+4f_4+5f_5+\cdots \\ &\geq 3(f_3+f_4+\dots)\\ &=3f.\end{align*}
Now apply Euler's formula to this inequality to get $e\leq 3v-6$. Your desired result follows immediately.
