Can the vertices of a planar graph of min degree 3 be covered with edges of average weight ( sum of degrees) at most 13?

Question

Loosely related to Every simple planar graph with $\delta\geq 3$ has an adjacent pair with $deg(u)+deg(v)\leq 13$ Since the proof is a kinda averaging argument, I wonder if one can get more than the existence of just one such edge. What is the smallest number that 13 can be replaced with so that the statement is true?

Answer 1

No.

Consider the counterexample to the linked question, the graph of a triakis icosahedron. This is a planar graph with 60 triangular faces, 12 vertices of degree 10, 20 vertices of degree 3, 60 edges of weight 13, and 30 edges of weight 20. It's possible to choose a cover from among the edges of weight 13, so this in itself isn't a counterexample.

Here is a planar drawing within an external triangle (from MathWorld):

Now take two copies of this and join the external vertices, like so:

Now every edge has weight at least 13, but any edge involving any of the six external vertices of the two triangles has weight at least 14. So any collection of edges covering every vertex must have average weight strictly greater than 13.