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Let $$d: \mathbb{R}^n \times \mathbb{R}^n \longrightarrow \mathbb{R}$$ be defined by $$d(x_i,x_j)=\frac{|x_i-x_j|}{\sqrt{M(i)M(j)}},$$ where $M(i)$ represents the average distance between $x_i$ and the other points, $M(j)$ represents the average distance between $x_j$ and the other points, and $|x_i-x_j|$ is the standard Euclidean distance. I need to prove that $d$ satisfies the triangle inequality. Thanks!

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This is false. The triangle inequality would be


Multiply through by $\sqrt{M(i)M(j)M(k)}$ to obtain


Now we can assume that there are an arbitrary number of points arbitrarily close to $x_k$, which would make $M(k)$ arbitrarily close to $0$, $M(i)$ arbitrarily close to $|x_i-x_k|$ and $M(j)$ arbitrarily close to $|x_j-x_k|$. Thus for the triangle equality to hold in all cases, we would need to have


and thus


which is obviously not always true.

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thank you, joriki. – user29860 Apr 26 '12 at 10:46

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