approximation algorithm for Travelling Salesman Problem: with a different inequality and not triangle inequality
Review the standard proof that it's a 2-approximation if the weight function satisfies the triangle inequality.  That suggests a way to upper-bound the total length of the Hamilton cycle $H$ returned by the algorithm.

Adjust that reasoning.  For starters, assume that every edge in $H$ is a "shortcut edge" $(u,w)$ that was taken by skipping a single vertex $v$.  What could you conclude about the cost of $H$, compared to the cost of $T$?  If you do a little bit of algebra you should be able to obtain an expression.

Now generalize to arbitrary $H$.