Trivial approximation ratio of n/2 for 2-Opt for TSPI recently read that the 2-Opt algorithm...

Step 1
For all those that stumble upon this question: The answer actually is quite simple. Let me explain.
Given an Instance $I=(K_n,w)$ of metric TSP, consider the following algorithm:
1. Construct an arbitrary Tour T through V.
2. Return T.
I claim that this simple algorithm already achieves an approximation ratio of n/2: Let $T^{\ast }$ denote an optimal Tour. Let $e=(v,w)$ be an arbitrary edge of T. Note that $T^{\ast }$ is the disjoint union of two paths $P_1$ and $P_2$ where one is a v-w-Path and the other is a w-v-Path. By the triangle inequality we have $w(e)\le w(P_1)=\sum _{f\in E(P_1)}w(f)$ and similarily $w(e)\le w(P_2)$. Together with $w(T^{\ast })=\sum _{f\in E(T^{\ast })}w(f)=w(P_1)+w(P_2)$ it follows that $2w(e)\le w(T^{\ast }).$
Step 2
Therefore $w(T)=\sum _{e\in E(T)}w(e)\le \sum _{e\in E(T)}\frac{1}{2}w(T^{\ast })=\frac{n}{2}w(T^{\ast }),$, which proves the claim.
My initial confusion came from the misunderstanding this approximation ratio would have something to do with the specific 2-Opt algorithm. However, as argued above, this is not the case and this approximation ratio is quite trivial to achieve.