Show that every planar graph admits an orientation such that each vertex has at most five outgoing edges An orientation of any graph $G = (V, E)$ is defined as any graph $G' = (V, E')$ arising by replacing each edge $\{u, v\}$ of $E(G)$ by a directed edge either  $\{u, v\}$ or $\{v, u\}$. Show that every planar graph admits an orientation such that each vertex has at most five outgoing edges. 
Intuitively I can see this to be true but I am having a hard time producing a formal proof for this. I was thinking of a case distinction where I can split the cases 1) the degree of any vertex of G is at most 5 (this case is trivially true) and 2) there exist vertices of degree more than 5 (having a lot of trouble trying to prove this). I'd really appreciate your help!
 A: Let $n$ be the number of vertices.

Proceed by induction on $n$.

For $n=1$, the result is trivially true.

Thus, suppose a planar graph $G$ has $n$ vertices, with $n \ge 2$.

But every planar graph has some vertex $v$ of degree at most $5$.

For a proof, see Theorem 5.8.6 (page 182), and Lemma 5.8.15 (page 184) of this MIT OCW textbook: 

https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/MIT6_042JF10_notes.pdf

Hence by the inductive hypothesis, $G-v$ can be oriented as specified.

To complete the orientation on $G$, just orient the edges from $v$ as outgoing.
A: Do you know the relatively easy proof of the six-colour and five-colour theorems for planar graphs? You can mirror the strategy for those as follows:


*

*Show that any planar graph has at least one vertex of degree at most five

*Consider a minimal counterexample. Deduct a vertex of degree at most five. Solve the smaller graph, and add the deleted vertex back in an appropriate way.

*Show that a graph with only one vertex and no edges has the relevant property (so the set of solvable graphs is not empty)
