Planar graph with $|V|\ge3$ has the average degree of vertices smaller than $6$ $G$ is a planar graph with $|V|\geq 3$. How to show that $G$ has an average degree of vertices smaller than 6?
Would it be enough to note that if $G$ has less then 6 nodes the average cannot be over 5 and the existence of connected components further decreases the $\frac{\sum_{v\in V}deg(v)}{|V|}$, so w.l.g. it can be proved for a connected graph (component) and extended to all connected components if needed?
Proof: If $G$ has no cycles it is a tree. $|E|=|V|-1$ and that is smaller than $|V|\leq |V|+2|V|-6 = |3|V-6$ so $$|E|\leq |3|V-6$$
If $G$ has cycles: we note that each face is bounded by 3 edges and the sum of the degrees of the faces has to be at least three times number of all faces: $$\sum_{f\in F}deg(f)\geq3|F|$$
Furthermore, each edge bounds at most two faces so $$\sum_{f\in F}deg(f)\leq2|E|$$
We would have $3|F|\leq2|E|$. Using Euler's formula we get $$3|F|=3|E|-3|V|-6$$ after multiplying it by scalar and combining both we get: $2|E|\geq 3|E|-3|V|-6$ and  $$|E|\leq 3|V|-6$$
Let's suppose that the degree of each node in $G$  is at least 6 (contraposition).
$$\frac{\sum_{v\in V}deg(v)}{|V|} \geq 6 \\  2|E|\geq 6|V| \\ |E|\geq 3|V|$$
But we have just shown that $|E|\leq 3|V|-6$, so $2|E|\leq 6|V|-12$. This two cannot be true at the same time for graph/connected component with $|V|\geq 6$.
If $G$ isn’t connected, the result still holds, because we can apply our proof to each connected component individually.
 A: Your proof is pretty much correct, and on the Wikipedia page for planar graphs , there is a brief explanation which is completely captured in your answer : first we show that $2|E| \geq 3 |F|$ , and then substitute in Euler's formula to get $|E| \leq 3|V|-6$. Then we just multiply by $\frac 2{|V|}$ on both sides, and rearrange to get that the average degree is at most $6 - \frac{12}{|V|}$, and therefore strictly less than $6$.
Now this is true for each connected component, but the average degree of the whole graph is at most the average degree of any connected component (using the pigeonhole principle), so the whole graph has average degree at most $6 - \frac {12}{|V|}$, where $|V|$ is the number of vertices.
Interestingly, we can study the average degree of a random planar graph. Although an upper bound on the degree is $6$, it was shown that if you uniformly sample, from all planar graphs with $n$ vertices, a random graph, then the number of edges of this graph is with high probability concentrated around $2.21n$ (at least for large enough $n$). This implies that the average degree of a uniformly randomly chosen planar graph is very highly concentrated around $4.42$, and $6$ is not really a very tight bound for a lot of planar graphs. On the other hand, planar graphs are also not trees, which have an average degree of at most $2$. So an average planar graph lies somewhere in the middle.
