Number of simple edge-disjoint paths needed to cover a planar graph

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Let $G=(V,E)$ be a graph with $|E|=m$ of a graph class $\mathcal{G}$. A path-cover $\mathcal{P}=\{P_1,\ldots,P_k\}$ is a partition of $E$ into edge-disjoint simple paths. The size of the cover is $\sigma(\mathcal{P})=k$. I am interested in upper and lower bounds for the quantity

$$ \max_{G\in \mathcal{G}} \quad \min_{\mathcal{P} \text{ is path-cover of $G$}} \sigma(\mathcal{P})/m.$$

In other words, how large is the inverse of the average path length in the smallest path-cover in the worst case.

The graphs classes $\mathcal{G}$ I am interested in are (1) planar 3-connected graphs, and (2) triangulations.

There is a simple observation for a lower bound: In every odd-degree vertex one path has to start/end. So when all vertices have odd degree there any path-cover needs at least $m/6+1$ paths (a triangulation has $3|V|-6$ edges). You get the same result by noticing, that $(n-1)/2$ paths have to pass through a vertex of degree $n-1$.

Is a path-cover with $\frac{m}{6}$ paths always possible for planar graphs?

I am aware of a few results covering planar graphs with paths of length $3$.

Here is an example of a path-cover of the graph of the icosahedron.

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