Equilibrium distance formula proof

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

$$\frac{|x_i-x_k|}{\sqrt{M(i)M(k)}}\le\frac{|x_i-x_j|}{\sqrt{M(i)M(j)}}+\frac{|x_j-x_k|}{\sqrt{M(j)M(k)}}\;.$$

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

$$|x_i-x_k|\sqrt{M(j)}\le|x_i-x_j|\sqrt{M(k)}+|x_j-x_k|\sqrt{M(i)}\;.$$

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

$$|x_i-x_k|\sqrt{|x_j-x_k|}\le|x_j-x_k|\sqrt{|x_i-x_k|}$$

and thus

$$|x_i-x_k|\le|x_j-x_k|\;,$$

which is obviously not always true.

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