I have a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ that is (strongly) convex (say in $\mathbb{R}^n$), but not necessarily differentiable. It attains its minimum at $\mathbf{q}$. Given two vectors $\mathbf{w},\mathbf{p}\in\mathbb{R}^n$, I would like to know if I can use the fact that $(\mathbf{w}-\mathbf{q})^\top(\mathbf{w}-\mathbf{p})\ge0$ to show that the directional derivative $f'(\mathbf{p};\mathbf{w}-\mathbf{p})$ is greater or equal than zero, where $f'(\mathbf{p};\mathbf{w}-\mathbf{p})=\text{inf}_{t>0}\frac{f(\mathbf{p}+t(\mathbf{w}-\mathbf{p}))-f(\mathbf{p})}{t}$. Intuitively, it should be (from my point of view), but I'm having a hard time to figure out how to show it. Any help would be appreciated.
Function $f$ is related to the function being minimized in a proximal operator, in case it matters.
Consider $f(x) = \frac12 x^2$ for $x\in\mathbb R$, $q=0$, $p=1$, $w = -1$. Then, we have $(w-q)(w-p) = -1(-2) = 2 \ge 0$ and $f'(1)(-2) = -2 < 0$.