$||x|| \le ||x+ry||$ for all $r \ge 0 \implies \langle j(x), y \rangle \ge 0$, where $j$ is the duality map.
Let $f(x) = \|x\|$. Then (1) implies $$ f'(x; y) \ge 0. $$ Now $f$ is a continuous convex function, $J(x)=\{j(x)\}$ is the subdifferential of $f$ at $x$, so $f'(x,y) = \langle j(x), y\rangle$.