Hardness of the Metric TSP for the Maximum Metric

I know that it is not too difficult to construct a metric to show that the metric TSP is NP-hard. The typical example is (1,2)-TSP. I also know that Papadimitriou has shown that Euclidean TSP is NP-hard. But I could not find anything about the hardness of the (2D) TSP with the maximum metric. Concretely, the problem is as follows:

Given a set of points $$P \subset \mathbb{R}^2$$, find a tour $$T = ((x_1, y_1), \dots, (x_{|P|}, y_{|P|}))$$ that contains each point $$p \in P$$ exactly once such that the tour cost $$\sum_{i=1}^{|P|-1} \max \{ |x_i - x_{i+1}|, |y_i - y_{i+1}| \}$$ is minimized.

Is this problem NP-hard?

From this, it follows that the 2D TSP with the Manhattan metric (i.e. $$dist = |x_1-x_2|+|y_1-y_2|$$) is NP-hard, as such graph has a Hamiltonian path if and only if we have such tour with cost $$n-1$$.
Now, to reduce the Manhattan metric to the max metric, we can simply use the mapping $$(x,y) \mapsto (x+y,x-y)$$. In particular, it holds that $$|x_1-x_2|+|y_1-y_2| = max(|x_1+y_1-x_2-y_2|, |x_1-y_1-x_2+y_2|)$$. This finishes the reduction.