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Reduce the Hamiltonian cycle decision problem to the metric travelling salesman problem. Given a simple graph on $n\ge 3$ vertices, form the complete graph with distance $1$ on original edges and distance $2$ on nonedges. These symmetric distances satisfy the triangle inequality, since a direct distance is at most $2$ and any two positive edge distances sum to at least $2$.

A tour has $n$ edges, each of cost at least $1$. It has cost at most $n$ exactly when every edge is an original graph edge, that is, exactly when the original graph has a Hamiltonian cycle. The construction and threshold are polynomial in the input size. Thus the metric travelling salesman problem is NP-hard; its rational-cost threshold version is also in NP because a tour is a polynomial-size certificate.

Allowing repeated vertex visits does not reduce the optimum for a complete metric instance. Given a closed walk visiting every vertex, keep the vertices in order of first appearance and shortcut the portions between them, including the final return. The triangle inequality makes the resulting Hamiltonian tour no more expensive. Conversely a Hamiltonian tour is an allowed closed walk. Hence the two optimal costs are equal, and the repeated-visit metric version remains NP-hard. If travel is described on a sparse graph, taking its shortest-path metric gives the corresponding closed-walk formulation rather than an easier exact problem.