How to prove that, if $G$ has an average node degree $\geq 6$ then $G$ cannot be planar

Question

Let $G= (V,E)$ simple, connected , undirected graph. Show that, if $G$ has an average node degree $\frac{1}{|V|}\sum_{v\in V} d(v) \geq 6$, then $G$ cannot be planar.

Answer 1

Hint. If $G$ is planar and connected, by Euler's formula, $|V|-|E|+|F|=2$. Now $G$ is simple and each face has at least three edges (provided there are at least three edges, see tomasz's comment and my P.S. below), therefore $3|F|\leq 2|E|$ and $$2=|V|-|E|+|F|\leq |V|-|E|+2|E|/3\implies |E|\leq 3|V|−6.$$ Finally use the fact that in any graph, $\sum_{v\in V}d(v)=2|E|$.

P.S. Note that $\frac{1}{|V|}\sum_{v\in V} d(v)=\frac{2|E|}{|V|} \geq 6$ implies that $|E|\geq 3$.