[proofplan]
We compare the minimum degree of $G$ with the average degree, using the handshaking lemma to express the latter in terms of the number of edges $e(G)$. The [Edge Bound for Planar Graphs](/theorems/2015) gives $e(G) \leq 3|G| - 6$, which when combined with $\sum_{v \in V(G)} \deg_G(v) = 2 e(G)$ yields an average degree strictly below $6$. Since the minimum of a set of integers is at most the average, $\delta(G) \leq 5$.
[/proofplan]
[step:Bound the total degree via the planar edge bound]
Let $G$ be a planar graph with $n := |G| \geq 3$. By the [Edge Bound for Planar Graphs](/theorems/2015), which applies to any simple planar graph on at least $3$ vertices,
\begin{align*}
e(G) \leq 3n - 6.
\end{align*}
Multiplying by $2$,
\begin{align*}
2 e(G) \leq 6n - 12.
\end{align*}
[guided]
Our goal is to bound $\delta(G)$ from above. The standard trick for minimum-degree bounds is to first control the average degree, then observe that the minimum cannot exceed the average.
To control the average degree, we need an estimate on $e(G)$ (the total number of edges), because the average degree is $\frac{2e(G)}{n}$ by the handshaking lemma. For planar graphs the standard ceiling on $e(G)$ is [Edge Bound for Planar Graphs](/theorems/2015), which requires $n \geq 3$ (the hypothesis of our theorem) and simplicity (which we assume throughout — the edge bound is stated for simple planar graphs).
The bound $e(G) \leq 3n - 6$ doubles to $2e(G) \leq 6n - 12$, setting up the handshaking lemma in the next step.
[/guided]
[/step]
[step:Apply the handshaking lemma to relate total degree to $e(G)$]
Recall the **handshaking lemma**: for any finite graph $H$,
\begin{align*}
\sum_{v \in V(H)} \deg_H(v) = 2 e(H),
\end{align*}
since each edge contributes exactly $2$ to the left-hand side (once for each endpoint). Applied to $G$,
\begin{align*}
\sum_{v \in V(G)} \deg_G(v) = 2 e(G) \leq 6n - 12.
\end{align*}
[guided]
The handshaking lemma is the combinatorial identity that turns edge counts into degree sums. It holds because in the double sum
\begin{align*}
\sum_{v \in V(G)} \deg_G(v) = \sum_{v \in V(G)} |\{e \in E(G) : v \in e\}| = \sum_{e \in E(G)} |\{v \in V(G) : v \in e\}| = \sum_{e \in E(G)} 2 = 2 e(G),
\end{align*}
each edge $e = uv$ is counted exactly twice — once at $u$ and once at $v$.
Combining with the planar bound from the previous step,
\begin{align*}
\sum_{v \in V(G)} \deg_G(v) \leq 6n - 12.
\end{align*}
This is the quantity that controls the average degree: divide both sides by $n$.
[/guided]
[/step]
[step:Compare the minimum degree to the average degree]
Since $\delta(G) \leq \deg_G(v)$ for every $v \in V(G)$, summing over $v$ yields
\begin{align*}
n \cdot \delta(G) = \sum_{v \in V(G)} \delta(G) \leq \sum_{v \in V(G)} \deg_G(v) \leq 6n - 12.
\end{align*}
Dividing by $n > 0$ (using $n \geq 3 \geq 1$),
\begin{align*}
\delta(G) \leq 6 - \frac{12}{n} < 6.
\end{align*}
Because $\delta(G)$ is a non-negative integer, a strict inequality $\delta(G) < 6$ forces $\delta(G) \leq 5$, completing the proof.
[guided]
The minimum of a finite collection of real numbers is at most their average: if $a_1, \ldots, a_n$ are real and $m = \min_i a_i$, then $nm \leq \sum_i a_i$, so $m \leq \frac{1}{n}\sum_i a_i$. Applying this to the degree sequence,
\begin{align*}
\delta(G) \leq \frac{1}{n} \sum_{v \in V(G)} \deg_G(v) = \frac{2 e(G)}{n} \leq \frac{6n - 12}{n} = 6 - \frac{12}{n}.
\end{align*}
Since $n \geq 3 \geq 1$, the fraction $\frac{12}{n}$ is strictly positive, so $6 - \frac{12}{n} < 6$. Hence $\delta(G) < 6$.
The final step is integrality: $\delta(G) \in \mathbb{Z}_{\geq 0}$ because it is the minimum of a set of non-negative integers (the degrees), so the strict inequality $\delta(G) < 6$ upgrades to $\delta(G) \leq 5$. This is the desired bound.
Note where the hypothesis $n \geq 3$ was used: it justifies invoking [Edge Bound for Planar Graphs](/theorems/2015) in Step 1, and also ensures $n > 0$ so we may divide. For $n = 1$ or $n = 2$ the statement $\delta(G) \leq 5$ remains true (and is trivial because $\delta(G) \leq n - 1 \leq 1$), but the proof via the planar edge bound does not apply.
[/guided]
[/step]
