TITLE: What makes a distance?
QUESTION [9 upvotes]: In the answers to my previous question, I learned that there are different concepts of distance, that is of distance-like functions with the usual metric being only the most popular and important one.
All distance-like functions seem to agree in being 

nonnegative: $d(x,y) \geq 0$
symmetric:  $d(x,y) = d(y,x)$
reflexive: $d(x,x)=0$ 

– let me call such functions proto-metrics – but differ in other respects:

a pseudo-metric fulfills the usual triangle inequality, but allows $d(x,y)=0$ for $x \neq y$.
an ultra-metric has $d(x,y)=0$ only for $x = y$, but fulfills a stronger triangle inequality, namely $d(x,z) \leq \text{max}(d(x,y),d(y,z))$ 
a dissimilarity function – conceived as some kind of distance, e.g. in a property space – is just any proto-metric, i.e. fulfills nothing but the core conditions above, esp. no triangle inequality at all.

One might ask whether such a dissimilarity function deserves its name, because it violates a strong intuition on dissimilarity (resp. distance) by allowing – without any restriction – two objects which are very similar (= close) to a third object to be arbitrarily dissimilar (= far away) from each other. 
Given that a distance-like function has to fulfill some kind of triangle inequality, my question boils down to:

Which conditions does a function $F$ have to
  fulfill such that
$$d(x,z) \leq F(d(x,y),d(y,z))\qquad(*)$$
can be seen as a reasonable triangle inequality,
  thus making the proto-metric $d$
  distance-like?

Some conditions on $F$ seem to spring to mind:

symmetry: $F(x,y)=F(y,x)$ 
monotonicity: $F(x,y) \leq F(x',y)$ for $x\leq x'$ 
scale-invariance: $\alpha F(x,y) \leq F(\alpha x, \alpha y)$ for $\alpha > 0$ 
associativity: $F(x,F(y,z)) = F(F(x,y),z)$ (Did I understand this correctly, Will Sawin? Where does this intution come from?) 
$F(x,0)=x$ (Thanks to David Feldman)

Is there hope to fix a set of conditions $\mathcal{C}$, such that one will be willing to accept any proto-metric $d$ to be distance-like if it fulfills $(*)$ for an $F$ that fulfills $\mathcal{C}$?

REPLY [3 votes]: Thinking about – and inspired by – an insightful example by Aaron Meyerowitz I found a convincing argument that there is probably no generally agreed upon concept of distance-likeness involving a sensibly generalized triangle inequality. The argument goes like this:
Consider a world of non-intersecting disks in the plane. Let the distance $d$ between two disks be the smallest Euclidean distance between any two points on their respective boundaries. This involves that any two touching disks have distance 0 and that the distance-law d(x,y)=0 iff x=y is violated. I assume that it is nevertheless generally agreed upon that this is a sensible distance. 
But any sensible triangle inequality is violated, too. That's because for every two non-touching disks $x,z$ that touch a common disk $y$ we have
$$d(x,z) > d(x,y) + d(y,z) = 0$$
and $d(x,z)$ can be arbitrarily large (depending on the radius of disk $y$) and thus greater than $F(0,0)$ for any function $F(u,v)$.
The question arises for what specific reasons we nevertheless do believe that $d$ is a
sensible distance?
Note, that for a given maximal disk radius $r$max we may get a sensible version of the triangle inequality (due to Aaron's example) and with $r$max $\rightarrow 0$ we get the usual triangle inequality.