Weighted Metric Graph: ratio of sum of wts of edges to the wt of MST
If the weight of the MST is $W$, then the weight of all edges is $\le \binom n 2 \cdot W$, because every edge in the graph is bounded by the unique path in the MST connecting the two nodes of the edge by the triangle inequality.

This is tight: Consider a line with edge weights all being equal to $1$ and let the metric be the shortest path metric induced by this line. Then the MST is this line and has weight $n-1$, on the other hand there are $\Omega(n^2)$ pairs of nodes with a distance of $\Omega(n)$.