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I'm trying to prove the following proposition.

Suppose $G = (V, E)$ is a planar graph such that each vertex in $V$ has degree at least five, and at least one vertex in $V$ has degree eight. Then, $G$ has at least 15 vertices.

My idea is to combine two bounds. The first comes from Euler's formula for planar graphs. Let $n = |V|$ and $e = |E|$. Then we have $e \leq 3n - 6$. The next bound comes from the assumption that every vertex has degree at least five and the Handshaking Lemma. $$ \sum_{v \in V} \operatorname{deg}(v) = 2 e \geq 5 n $$ This doesn't use the assumption that there is at least one vertex with degree eight. It's a bit unclear to me how I can bump up the bound with that assumption though.

I think I can just say, $2e \geq 5 n + 3$ because there are at least three vertices that are unaccounted for by my use of the handshaking lemma.

(The rest is easy. I combine the bounds by transitivity to get $2(3n - 6) \geq 5n + 3$. Some algebraic manipulation gives $n \geq 15$ as required.)

Could someone confirm the step where I bump up the lower bound? Thanks in advance.

1 Answers 1

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You are right everywhere except assumption about three magic vertices.

Let $u \in V(G)$ be a vertex of degree 8. Then $$2e = \sum_{v \in V(G)} \deg v = \deg u + \sum_{v \in V(G) \setminus \{\,u\,\}} \deg v \ge 8 + 5(n - 1) = 5n + 3.$$