The Prize-Collecting TSP (PCTSP) is defined as the ordinary TSP with the difference that penalties are added to nodes; so we may avoid visiting a node paying its penalty, which is added to the overall cost. It currently admits a $2-\epsilon$ approximation ratio, where edge costs are considered to satisfy the triangle inequality. Obviously, ordinary TSP is PCTSP where all node penalties are set to inf.
I would like to ask, in the case a graph which does not satisfy the triangle inequality is chosen and ALL of its edges are deleted, replaced by 2 new edges connected to a 2-degree new node for each case of deleted edge, where each edge cost is equal to the deleted edge's cost/2 and having Euclidean distance equal to that (so the new graph is metric), don't we get a $(2-\epsilon)$ approximation ratio for the original (general and therefore non-approximable) graph? (Penalties for old nodes are set to inf and to new 2-degree nodes equal to 0).
Of course, there is a mistake in my thought, since this can't be true, so if somebody finds it, I would be pleased to listen to that. Thank you in priori.