The full version of this question is quoted below:
Let G be a connected graph with n vertices, m edges with distinct edge weights. Let T be a tree of G with n vertices and n-1 edges (i.e. a spanning tree), and define a bottleneck edge of T to be the edge of T with the smallest weight. The max-bottleneck tree is a spanning tree of G if there is no spanning tree with larger bottleneck edge. Prove or provide a counter example for the following statement:
Every max-spanning tree of G is a max bottleneck tree of G
I think since the graph has unique edge weights, then every spanning tree of G is also unique. Then there is only one maximum spanning tree of G, and if I can prove that this tree is also a max bottle neck tree, then that would prove this statement to be true, but only if it's true for all graphs that have unique edge weights.
I've tried looking for counter examples to prove this false but so far it looks like every graph I draw with unique edge weights winds up have the max spanning tree also be a max bottleneck tree. I think I can use that to prove that this statement is true, but I am not sure how to word it.
Negate all the edge weights in the graph. Then the problems get changed to Minimum Spanning Tree and Minimum Bottleneck Spanning Tree respectively.
Now every Minimum Spanning Tree is also a Minimum Bottleneck Spanning Tree. Proof by Cut Property.
http://flashing-thoughts.blogspot.in/2010/06/everything-about-bottleneck-spanning.html