There are a number of good answers from differential geometry, analysis, etc. But I suspect the questioner had axiomatic geometries in mind. So here's another take from that perspective.

There are two statements of the triangle inequality in plane geometry.
(1) If A,B,C are noncollinear points, then $AC\lt AB+BC$;
(2) If A,B,C are any three points, then $AC\leq AB+BC$.

In any system which includes a Ruler Postulate, (1) is stronger than (2).
In neutral geometry (which includes all of the axioms of Euclidean geometry except the parallel postulate), statement (1) can be proven pretty directly from the SAS congruence postulate. (Does anyone know if they're equivalent in the presence of the other neutral geometry axioms?) Statement (2), on the other hand, is the one needed to show that the plane is a metric space, and is strictly weaker, as shown by the "taxicab geometry" mentioned above, which satisfies all of the postulates for neutral geometry except for SAS, and has property (2) but not property (1).

So, to summarize, the triangle inequality is true in neutral geometry, so any model of it (including the Euclidean and hyperbolic planes, etc.) will satisfy the triangle inequality. But of course we can consider weaker axiom systems in which models do not satisfy it (like taxicab).