Question on the Prize-Collecting TSP's ratio related to inapprox. of general TSP 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.
Edited may answer, I hope it becomes clearer now (or I find out what I did not understand):

Your process is as follows: If you have an edge $e = \{u,v\}$ with weight $t$ you replace $e$ by a path of length two with edges $\{u,z_e\}$ and $\{z_e,v\}$. $z_e$ is a new node and each of the two new edges get weight $t/2$. For each edge, we introduce a new node.

So far, we do not have a metric instance, because there is no edge between $u$ and $v$ so far and the distance between $u$ and $v$ is infinite. Therefore, you take the shortest path metric to get these distances. (I suppose that this is what you mean by "having Euclidean distances equal to that". Otherwise, you should clarify this.)

My point is that after adding distances according to the shortest path metric, the distance between $u$ and $v$ might not be $t$ anymore, so you do not get your original instance back:
Consider a triangle with edge lengths $1$, $1$ and $t$. Let $u$ and $v$ be the nodes of the edge with weight $t$. Apply your construction. The shortest path between $u$ and $v$ is $2$ (it takes the four edges with weight $1/2$) and not $t$.

Btw: Euclidean distance should just be distance in the question.