Does convexity of a 'norm' imply the triangle inequality?
Solution 1:
Resolved in comments.
Setting $p=\frac12$ in the definition of convexity, we have $$ d\Big( \frac{\mathbf u + \mathbf v}{2} \Big) \leqslant \frac12 d(\mathbf u) + \frac12 d(\mathbf v). $$ By the scaling or homogeneity, the left hand side is simply $\frac12 d(\mathbf u + \mathbf v)$; plugging in this and simplifying, we get the triangle inequality.