Past exam of the short options course of the University of Oxford / 2023 / 7 / b / iii / Solution

Take a minimum spanning tree $T$, for example one constructed by Kruskal's algorithm. Fix $v$ and let $e$ be a longest edge of the tree's unique $xv$ path. Deleting $e$ partitions $T$ into two components separating $x$ and $v$. Any alternative $xv$ path crosses this graph cut. If all of its edges were shorter than $e$, a crossing edge $f$ would let us replace $e$ by $f$, producing a spanning tree of smaller total length. This contradicts minimality of $T$.

Therefore every competing path has an edge at least as long as $e$, and the tree path solves the bottleneck shortest path problem. The same minimum spanning tree works simultaneously for every $v$; this is the minimax path property of a minimum spanning tree.