Every planar graph is a union of $3$ forests I am currently working on problems related to planarity in graphs, and
I came across a peculiar problem related to proving the relationship between
planar graphs and forests.
Take a planar graph $P$. I imagine that I want to induce $P$ to satisfy
Nash-Williams' theorem. It was a suggestion to me that I consider proving that
a planar graph has at most $3n-6$ edges, which would by Nash-Williams imply
that a planar graph can be partitioned into $\frac{3n-6}{n-1} = 3$ forests.
I know that a planar graph satisfies $v-e+f = 2$ where $v$ is the number of
vertices, $e$ is the number of edges, and $f$ is the number of faces. If I can
contradict the statement that $e > 3v-6$, I may be able to conclude this proof.
We see that if we assume for the sake of contradiction $e > 3v-6$, we see that
$$-e < -3v + 6 \implies v-e < -2v + 6 \implies v-e+f < -2v+f+6,$$
which by planarity implies that $-2v +f + 6 = 3.$ However, I am having issues
on where to move here. Is there a contradiction found in this condition?
 A: Your proof may be on the wrong track, since you can only utilise Euler's formula if the graph is connected, and I cannot think of how to make that work out. I provide a proof below which uses triangulations, which are connected and Euler's formula will apply.
This inequality is well known. For $n=1$ and $n=2$ it is obvious. First note that every planar graph with $v\ge 3$ vertices is a subgraph of a plane triangulation with the same vertex set (a planar graph where all faces have degree $3$). To see this, for every face with degree larger that $3$, we can choose one vertex $u$ on its boundary and add new edges from $u$ to every nonadjacent vertex on the boundary of the face; the resulting graph is a triangulation.
Let $e^\prime$ be the number of edges in the triangulation. Here, there are $3$ edges per face, so $3f=2e^\prime$ (each edge is adjacent to two faces). Then by Euler's formula, $$3e^\prime=3(v+f-2)=3v+2e^\prime-6.$$
Since your graph is a subgraph of the triangulation, $e\le e^\prime$; hence, after rearranging,
$$e\le 3v-6.$$
