Every simple planar graph with $n\ge3$ vertices has at most $3n-6$ edges, with equality for every plane triangulation

First suppose the graph is connected. If it is a tree then $m=n-1\le3n-6$. Otherwise simplicity makes every facial boundary walk have length at least three, so [L2] gives $3|F|\le2m$. Combining this with [L1], $|F|=2-n+m$, yields $3(2-n+m)\le2m$, hence $m\le3n-6$.

[L2] supplies more than its inequality: for a connected plane graph it gives the exact facial total $\sum_{f\in F}\ell(f)=2m$, from which that inequality is only the consequence drawn there.

If the graph is disconnected, first redraw it. Each component drawing is a finite union of segments and so is bounded, so translating and scaling the components into pairwise disjoint discs gives a plane drawing of the same abstract graph in which every component meets the unbounded face; the bound to be proved does not depend on the drawing. Now join the components by noncrossing edges through that face until the drawing is connected. This preserves simplicity and planarity, keeps $n$ fixed, and only increases the number of edges. Step 1.1 applied to the augmented graph therefore bounds the original $m$ by $3n-6$.

Let the graph be a plane triangulation with $n\ge3$ vertices. By [L3] it is connected, and every face has boundary a triangle, so $\ell(f)=3$ for every face. Then step 1.2 reads $3|F|=2m$, and [L1] gives $|F|=2-n+m$, so $3(2-n+m)=2m$ and $m=3n-6$.