Any reasonable notion of distance satisfies the triangle inequality, since if you can get from point $A$ to point $B$ using a route of length $d(A, B)$ and from point $B$ to point $C$ using a route of length $d(B, C)$, you can clearly get from point $A$ to point $C$ using a route of length $d(A, B) + d(B, C)$, and the optimal route (of length $d(A, C)$) therefore can't be any longer than this. (This argument is completely rigorous in an intrinsic metric, but there's no reason not to take it as motivation in general.)
Lawvere noticed that the above argument looks an awful lot like composition of morphisms in a category, and that led him to the categorical definition of metric spaces as certain kinds of enriched categories. From this perspective neither the symmetry nor the positive-definiteness axioms are particularly natural, but the triangle inequality is still very natural.
@Rasmus: letting $x = y$ in the reverse triangle inequality gives $0 \ge 2d(x, z)$, so the only "reverse metric space" is a point.