Call the differentiable function $f: \mathbb K \to \mathbb R$ on some compact convex $K \subset \mathbb R^n$strictly convex to mean $f(\lambda x + (1- \lambda)y) < \lambda f(x) + (1-\lambda)f(y)$ for all $x,y \in K$ and $\lambda \in (0,1)$.
For $n = 1$ strict convexity is equivalent to the derivative $f'$ being strictly increasing. In that case we can show $\min f$ occurs at the minimiser of $|f'|$.
Consider first the case that $f'(a) = 0$ for some $a \in K$. By strict convexity there is only one such $a$. Clearly $a$ is the unique minimiser of $|f'|$. But $f'(a) = 0$ also means $a$ is a local minimum, and then convexity implies $a$ is the global minimum.
Otherwise $f'$ is strictly positive or negative negative. First assume the former. Without loss of generality we have $K =[0,1]$. Then since $f'$ is strictly increasing $|f'(x)| = f'(x)$ has unique minimum at $0$. Also by writing $f(x) = f(0) + \int_0^x f'(y) \, dy$ as the integral of a strictly positive function we see $f$ is also strictly increasing hence also has unique minimum at $0$. A symmetric argument shows for $f'$ strictly negative both functions have unique minimum at $1$.
Of course for $n>1$ the derivative of $f$ is the gradient vector $\nabla f$ of partial derivatives. Since there is no notion of a vector being positive the proof fails to generalise. Also the higher-dimensional analogue of $f(x) = \int_0^x f'(y) \, dy$ is Stoke's theorem which does not recover values of $f$ at a point. Rather the left-hand-side becomes the integral over the boundary of whatever region we're integrating over on the right.
Nevertheless my intuition says some variant of the result above should hold in higher dimensions. Does anyone have any ideas or a good reference?