Bottleneck TSP with MST
You need to prove by induction the following claim:


  Let $T$ be a tree rooted at $r$ with at least two vertices. You can list the vertices of $T$ in a sequence such that the distance (in $T$) between two adjacent vertices (including the first and the last) is at most $3$, and furthermore the vertex following $r$ is adjacent to $r$ (in $T$).


(The induction is on the number of vertices.)

Given this, you argue as follows. Suppose that there is a Hamiltonian cycle with heaviest edge having weight $w$. Then there is a spanning tree whose heaviest edge has weight $w$. So the heaviest edge in the bottleneck spanning tree has weight at most $w$. Apply the claim to that tree, and consider the corresponding Hamiltonian cycle. Since the tree distance between adjacent vertices is at most $3$ and thanks to the triangle inequality, the heaviest edge in the resulting cycles weighs at most $3w$.