 Solution

Let $T$ be the unique minimum spanning tree. If $e=ab$ is an edge outside $T$, adding it creates a unique graph cycle consisting of $e$ and the tree graph path $P_T(a,b)$. Every edge $f$ on that graph path has $w(f)<w(e)$: otherwise the distinct weights would give $w(f)>w(e)$, and replacing $f$ by $e$ would produce a cheaper spanning tree. This is the minimum spanning tree cycle property in exchange form.

By the domination inequality in (i), it follows more strongly thatNow consider any simple graph path $P$ from $v$ to $u$ that is not the tree graph path. It must use an edge outside $T$. Replace every such edge by the corresponding tree graph path. Each replacement strictly reduces length and produces a walk entirely in $T$. Removing any repeated-vertex excursions leaves the unique tree graph path $P_T(v,u)$ and can only reduce length further. ThereforeSo every root-to-vertex graph path in $T$ is the unique shortest graph path in $G$, andThe argument works for every choice of root $v$.