This is just an elaboration on Brendan McKay's beautiful answer, but too long for a comment. The crucial idea is to simplify the problem by generalising it, introducing a maximisation on the indices of the sum, detached from the original planar graph $G$.

For $x = (x_1, \ldots, x_n) \in \mathbb{R}_{\geq 0}^n$ and a multi-graph $H$ with $V(H) \subseteq \{ 1, \ldots, n \}$, let
$$
L_H(x) := \sum_{ij \in E(H)} \min \{ x_i, x_j \} .
$$
For a natural number $d$, let $\mathcal{H}_d$ be the class of all finite multi-graphs $H$ with $V(H) = \{ 1, \ldots, n \}$ (for any $n$) such that $e(H[A]) \leq d(|A|-1)$ holds for any $A \subseteq V(H)$ (in other words, graphs of arboricity at most $d$).

*Theorem:* For any $x \in \mathbb{R}_{\geq 0}^n$ and any $H \in \mathcal{H}_d$ we have $L_H(x) \leq d(\sum_{i=1}^n x_i - \max_i x_i)$.

*Proof:* We may assume that $x_i >0$ holds for every $i$. Permute $\{1, \ldots, n\}$ so that $x_1 \geq x_2 \geq \ldots \geq x_n$. Note that
$$
L_H(x) = \sum_{ij \in E(H) \atop i \, < \, j} x_j
$$

Extend the class $\mathcal{H}_d$ slightly by requiring only that $e(H[A]) \leq d(|A| - 1)$ holds when $A = \{ 1, \ldots, k \}$ for some $k$ (that is, $A$ is an initial segment of $\{ 1, \ldots, n\}$). Call this new class $\mathcal{H}_d'$.

Choose $H \in \mathcal{H}_d'$ with $V(H) = \{ 1, \ldots, n \}$ so that $L_H(x)$ is maximum and, subject to this, so that
$$ R(H) := \sum_{ij \in E(H)} i + j$$
is minimum. We claim that $H$ is a star with center 1.
Indeed, if $ij \in E(H)$ and $1 < i < j \leq n$, then we could replace $ij$ by the edge $1j$ and obtain a graph $H'$ with $L(H') = L(H)$ and $R(H') < R(H)$. Trivially $H' \in \mathcal{H}_d'$ as well, hence contradicting our choice of $H$.

Moreover, every $j \in \{ 2, \ldots, n \}$ has degree exactly $d$ in $H$. Otherwise, choose $j$ minimum with $d_H(j) \neq d$. If $d_H(j) > d$, then for $A := \{ 1, \ldots, j \}$ we have $e(H[A]) > d(|A| - 1)$, a contradiction to $H \in \mathcal{H}_d'$. Hence $d_H(j) \leq d-1$. Add an edge $1j$ to $H$. If there is some $k > j$ with $d_H(k) > d$, take a minimum such $k$ and delete one edge $1k$. If there was no such $k$, then the resulting graph $H'$ satisfies $L(H') > L(H)$. If there was such a $k$, then $L(H') = L(H)$ and $R(H') < R(H)$. Either way, $H'$ is easily seen to lie in $\mathcal{H}_d'$ and thus contradicts our choice of $H$.

Now that $H$ is explicitly given, it follows that
$$
L_H(x) = \sum_{j = 2}^n d x_j .
$$
This finishes our proof.

The original statement now follows by taking as $x$ the degree-sequence of a planar graph and noting that $G \in \mathcal{H}_d$ for $d = 3$.

averages of higher order: it seems that theaverage of the average of your function(the number defined above) can more or less routinely be calculated to a precision you perhaps did not expect. For example, it seems that said second-order-average can be expressed "in terms" of elementary functions, and hence numerical approximation can be calculated to very many digits of accuracy. (But: somethingmore complicatedthan "elementary functions" in the contemporary technical sense is involved.) $\endgroup$ – Peter Heinig Jul 8 '17 at 13:36the average of the average of your functioninvolves something like afunction elementary relative to a non-algebraic equation. $\endgroup$ – Peter Heinig Jul 8 '17 at 13:44