Is there a chance to make a sound argument for the triangle inequality - characterizing distances - from general considerations only, e.g. like this:

Given arbitrary distances $d(x,y)$ and $d(y,z)$ the distance $d(x,z)$ cannot be greater than a specific value which depends on $d(x,y)$ and $d(y,z)$. That is

$$d(x,z) \leq F(d(x,y),d(y,z))$$

Because $F$ must be symmetric, scale-invariant, and - by Occam's razor - as simple as possible, $F(a,b)$ must be $a + b$.

Is the last step (a) valid, and if not so (b) why?

I ask this question thinking of *closeness* $c(x,y) \in [0,\infty)$ as a natural counter-concept to *distance*. You can measure closeness just as you can measure distance (by applying yardsticks correspondingly). Closeness is symmetric, and closeness is maximal (as opposed to minimal) iff $x=y$. All in all: closeness is obviously nothing but and definable as the reciprocal of distance. **BUT**: one might want to establish it on its own feet. And this would demand of a reciprocal triangle inequality:

Given arbitrary closenesses $c(x,y)$ and $c(y,z)$ the closeness $c(x,z)$ cannot be smaller than a specific value which depends on $c(x,y)$ and $c(y,z)$. That is

$$c(x,z) \geq G(c(x,y),c(y,z))$$

Because $G$ must be symmetric, scale-invariant, and - by Occam's razor - as simple as possible, $G(a,b)$ must be $\frac{ab}{a+b}$.

Questions:

Does the last step here disturb you more than in the first case?

If so: why?

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