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}

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}

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}

Now the left hand side has a lower bound of 0. Because $x^{\top}x \geq 0$.