The same reasoning for the upper bound can be applied for the lower bound. Which was:
\begin{equation} (f(\xi)-f(\tilde{\xi}))^{\top} \leq (\xi - \tilde{\xi})^{\top} (\frac{\partial f}{\partial \xi} (\bar{\xi}))^{\top} \end{equation}\begin{equation} \left(f(\xi)-f(\tilde{\xi})\right)^{\top} \leq (\xi - \tilde{\xi})^{\top} \left(\frac{\partial f}{\partial \xi} (\bar{\xi})\right)^{\top} \end{equation}
Such that taking the transpose of this inequality gives:
\begin{equation} (f(\xi)-f(\tilde{\xi})) \leq (\frac{\partial f}{\partial \xi} (\bar{\xi})) (\xi - \tilde{\xi}) \end{equation}\begin{equation} \left(f(\xi)-f(\tilde{\xi})\right) \leq \left(\frac{\partial f}{\partial \xi} (\bar{\xi})\right) (\xi - \tilde{\xi}) \end{equation}
Filling this into the left hand side of the original inequality gives:
\begin{equation} 0 \leq (f(\xi)-f(\tilde{\xi}))^{\top} (f(\xi)-f(\tilde{\xi})) \leq (f(\xi)-f(\tilde{\xi}))^{\top} (\frac{\partial f}{\partial \xi} (\bar{\xi})) (\xi - \tilde{\xi}) \end{equation}\begin{equation} 0 \leq \left(f(\xi)-f(\tilde{\xi})\right)^{\top} \left(f(\xi)-f(\tilde{\xi})\right) \leq \left(f(\xi)-f(\tilde{\xi})\right)^{\top} \left(\frac{\partial f}{\partial \xi} (\bar{\xi})\right) (\xi - \tilde{\xi}) \end{equation}
Now the left hand side has a lower bound of 0. Because $x^{\top}x \geq 0$.