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

A minnimum-spanning-tree (MST) path is always V-1 number of edges and Hamiltonian Cycle (HC) always V number of edges. Because the HC has an extra edge we could say that in general, the weight of every Hamiltonian cycle of a connected graph will be less.

If we take an edge of every Hamiltonian cycle we will find an MST and by that, the MST weight will be always lower than HC.

  • There is a way to prove this where MST tree not necessary belongs to the HC?
  • There is more definitive proof for this?