Proved the hardness results for exact metric TSP already (since 
).
The hardness for a ratio of 2 come from replacing the distance 2 edges with distance 2 n edges. Then, if there is a Hamiltonian circuit, the approximation algorithm would have to give us a tour of length no more than 2 n. Such a tour could only consist of the length 1 edges and, therefore, would indicate the existence of a Hamiltonian circuit in the original graph.
