Relationship between $\langle \nabla f(\overline{x}), x - \overline{x}\rangle > 0$ and the the minimality of $\overline{x}$. Say $f$ is a differentiable function over a convex set $X$ with $\langle \nabla f(\overline{x}), x - \overline{x}\rangle > 0$ for all $x, \overline{x} \in X$ such that $x \neq \overline{x}$. Can we conclude from this alone that $\overline{x}$ is a local minimum? Certainly, if $f$ were pseudoconvex, this would hold trivially. However, I suspect that the strict inequality here may imply that $\overline{x}$ is indeed a local minimum

Assuming you meant that $\langle \nabla f(\bar{x}), x - \bar{x}\rangle > 0, \forall x \neq \bar{x}$, this should imply a local minimum. From the differentiability of $f$, you know that near $\bar{x}$:
$$
f(x) = f(\bar{x}) + \langle \nabla f(\bar{x}), x - \bar{x} \rangle + o(\| x - \bar{x}\|),
$$
where $o(\| x - \bar{x} \|)$ is a term that satisfies
$
\lim_{x \to \bar{x}} \frac{o(\| x-\bar{x}\|)}{\| x - \bar{x}\|} = 0.
$
Then, assuming that $\forall \epsilon > 0$, there is a $x$ with $f(x) < f(\bar{x})$ and $\| x - \bar{x} \| < \epsilon$ (i.e. $\bar{x}$ is not a local minimum), you should
be able to derive a contradiction for a sufficiently small $\epsilon$.
