# An inequality about median of points in higher dimensions

Let $$S$$ be a set of points in $$\mathbf{R^d}$$ and let $$m$$ be the median of this set of points, i.e. $$\sum_{x \in S} || x - y||$$ is minimized when we have $$y=m$$. Now let $$z$$ be an arbitrary point in $$\mathbf{R^d}$$. Is the following true?

$$\sum_{x \in S} ||x-z|| \leq K \cdot (\sum_{x \in S} ||x-m|| + |S| \cdot ||m-z||)$$ where $$K$$ is a constant.

We know that if we replace all the $$||a-b||$$ expressions by $$||a-b||^2$$, and we consider the mean instead of the median(i.e. replace $$m$$ with $$\mu$$), then actually we will have equality and $$K=1$$. So I want to know if similar things hold for the median.

I was thinking maybe the following is true. Is it?

$$\sum_{x \in S} ||x-z|| \leq \cdot (\sum_{x \in S} ||x-m|| + 4|S| \cdot ||m-z||)$$

• Doesn't triangle inequality give you $\| x - z\| \le \|x - m\| + \|m - z\|$? In which case the inequality holds even with $K=1$ and replacing "$|S|$" with 1. And for arbitrary $m$? What am I missing? Commented Feb 16, 2023 at 11:49
• @NealYoung The $|S|$ thing is still needed, as you are summing $|S|$ triangle inequalities together. Commented Feb 16, 2023 at 11:58
• Oh, I thought that term was inside the summand. It makes more sense that it is not. Commented Feb 16, 2023 at 17:03

Yes. By the triangle inequality, $$\|x-z\| \le \|x-m\| + \|m -z\|$$, which implies the desired inequality (with $$K=1$$) for any $$m$$ and $$z$$.