$||x|| \le ||x+ry||$ for all $r \ge 0 \implies \langle j(x), y \rangle \ge 0$, where $j$ is the duality map. Let $X$ be a real Banach space. Let $J \colon X \to 2^{X^*}$ be its (normalized)  duality map,
$$ J(x) = \{ x^* \in X^* \colon \langle x^* , x \rangle =||x|| \ ||x^*||, \ || x^* ||=||x||   \} , \ x \in X.$$
Assume that $X$ is smooth, so that $J(x)= \{j(x)\}$ is a singleton. Fix $x,y \in X$. It is known that, if
$$||x|| \le ||x+ry||,  \tag 1$$ for every $r \in \mathbb R$, then $\langle j(x),y \rangle =0$.   What happens if we only take $r \ge0$ in $(1)$? I expect that $ \langle j(x),y \rangle  \ge 0$, since that is the case for Hilbert spaces. Indeed, if $X$ is a Hilbert space with inner product $(\cdot,\cdot) $, then $J$ is just the identity map, and $(1)$ implies that
$$ r^2 ||y||^2 + 2r (x,y) \ge 0, $$
for every $r \ge 0$.  Dividing by $r$ and then letting $r \to 0$ we obtain that $(x,y) \ge 0$.
 A: 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 $f'(x,y) = \langle j(x), y\rangle$.
A: Here is a different approach. For $r >0$ consider the unit functional $y^*_r = j(x+r^{-1}y)/\|j(x+r^{-1}y)\|$. Then
\begin{align*}
\|x\| &\le \|x+r^{-1}y\| =  \frac{1} {\| j(x+r^{-1}y) \|}\langle j(x+r^{-1}y),x+r^{-1}y\rangle  = \langle y^*_r, x +r^{-1}y\rangle 
\\
&=\langle y_r^* ,x \rangle + \tfrac 1r\langle y_r^* ,y\rangle \le \|x\| + \tfrac 1r \langle y_r^* ,y\rangle.
\end{align*}
Since the closed unit ball of $X^*$ is weak-star compact, $(y_r)_{r>0}$ admits a weak-star convergent subnet. If $y^*$ is its limit point, then it is true that
$$  \|y^*\| \le 1, ~~~ \langle y^*,y\rangle \ge 0 ~~~\text{and} ~~~\langle y^*,x\rangle \ge \|x\|.$$
Since $\langle y^*,x\rangle \le \|x\|,$ we obtain that
$\langle y^*,x \rangle=\|x\|$ and thus $\|y^*\|=1$. It is easy to see that $j(x) = \|x\|y^*$ and so
$$\langle j(x),y\rangle= \|x\| \langle y^*,x \rangle \ge 0.$$

Note: The converse is also true. That is, if $\langle j(x),y)\ge 0$ then for any $r>0$
$$ \|x\| \le \|x+ry\|.$$
The proof is similar to the case of  Hilbert spaces: We estimate
$$ \|x\|^2 = \langle j(x),x \rangle  \le \langle j(x), x \rangle +r \langle j(x),y \rangle = \langle j(x) , x+ry\rangle  \le \|x\| \|x+ry\|.$$

