Every planar graph has a vertex of degree at most 5. I am trying to prove the following statement, any help!?

Prove that every planar graph has a vertex of degree at most 5. 

 A: Assume that every vertex of a planar graph $G$ has degree $6$ or more. Thus by The First Theorem of Graph Theory we have $2m\geq 6n$ where $n$ and $m$ denote the order and size of $G$. Since $G$ is planar we know that $m\leq3n-6$ and we have $3n\leq m$. Thus $3n\leq 3n-6$ which implies that $0\leq -6$ which is a contradiction. Thus every planar graph has a vertex of degree at most $5$.
A: Prove it by contradiction. Suppose that there exists a planar graph with all vertices having degree at least $6$. Then by the handshake lemma, $E = 3V$. If $G$ is planar, then $E \leq 3V - 6$, a contradiction.
A: For the sake of contradiction, suppose that $\exists$ a connected, planar graph, $G = (V, E)$, such that for all $v \in V(G)$, $deg_G(v) > 5$. We know that $$\displaystyle\sum_{v \in V(G)} deg_G(v) = 2|E| \leq 6|V| - 12 $$ must hold because $\displaystyle\sum_{v \in V(G)} deg_G(v) = 2|E|$ and $|E| \leq 3|V| - 6$ (if you have not see these formulas before, I would suggest trying to prove them on your own or at least understanding their respective proofs). Also, because we've assume that for all $v \in V(G), deg_G(v) \geq 6$, we have $$2|E| = \displaystyle\sum_{v \in V(G)} deg_G(v) \geq 6 \cdot |V| > 6 \cdot |V| - 12$$. Notice that we have $$2|E| > 6|V| - 12$$, which is a direct violation of the rule $|E| \leq 3|V| - 6$. Hence, a contradiction.
