Does a Minimum-Spanning-Tree always give a lower bound for the weight of any Hamiltonian cycle of the graph?
Suppose:


$G$ is some graph with non-zero, positive weights.
The minimum spanning tree of $G$ is denoted by $MST$
$H_C$ is a hamiltonian cycle in $G$.
$H_P$ is a hamiltonian path in $G$, given by removing any one edge of $H_C$.


Since all the edges are positive and non-zero, we immediately get that: 

$$w(H_C) &gt; w(H_P)$$

Also, $H_P$ is a subgraph that includes all vertices of $G$ (and connects them), so $H_P$ in fact spans $G$. 

Hence, by definition of minimum spanning tree:

$$w(H_P) \geq w(MST)$$

It follows that $$w(H_C) &gt; w(H_P) \geq w(MST)$$



In case we allow zero weights, the inequality holds, it simply will not be strictly greater (for the same reasons noted in the answer). That is:
$$w(H_C) \geq w(H_P) \geq w(MST)$$