I want to prove the following claim for undirected and connected graph, Spanning Tree (T) and weight function with real numbers:
$T$ is a minimum spanning tree in $G=(V,E)$ iff for every edge $e=u-v$ in $E$ the weight of it is bigger or equal to the heaviest edge in the only (simple) path between $u$ and $v$ in $T$
If $T$ is a minimum spanning tree in $G=(V,E)$ it's easy to prove the claim, but what about the other direction?
I've no clue, plus I tried to assume that $T$ is NOT a minimum spanning tree in $G=(V,E)$ and try to reach a contradiction, but that assumption doesn't seem to add much of information (if a spanning tree isn't MST then what can I see more than that)?
Let T satisfy the condition, let us prove that it is a MST.
By contradiction, if it is not there is a different MST S. Let (u,v) be in S and not in T, if we remove it from S we are left with a graph S' with exactly two connected components. Now observe (hint : T is spanning) there is at least one edge (a,b) != (u,v) in T that connects them. By the condition, (a,b) has a smaller weight than (u,v), so if we add it to S' we obtain a spanning tree with smaller weight than S, which is absurd since S was supposed to be a MST.
Conclusion : T is a MST.