I need to prove that in a planar graph with more than two vertexes at least a vertex with a maximum grade of five exists.
I know that planar graphs fulfill the relation $\#E \le 3\#V-6$, and relation $\#E \le 2\#V-4$ if there is no subgraph that is isomorphic to $K_{3}$. $\#E$ being the number of edges and $\#V = n$ being the number of vertexes.
Based on the statement that says $\sum_{i=1}^{n}gr(v_{i}) = 2\#E$ I have come up that for complete graphs $K_{n}$ the next relation must be fulfilled: $$n^{2}-4n+6\le 0$$ But I can not figure how to prove what it is asked to prove
Suppose all vertices have degree 6 or more. Then we have $$ |E|\geq \frac{1}{2}\cdot 6 \cdot|V|=3|V|, $$ since every edge contains two vertices. This clearly contradicts your inequality.