Suppose $\mathbf{D} = \begin{bmatrix} \textbf{A} & \textbf{b} \\ \textbf{b}^T & c \end{bmatrix}$ where $\textbf{A} \in \mathbb{R}^{n \times n}$, $\textbf{b} \in \mathbb{R}^n$ and $c \in \mathbb{R}$. Suppose also that $\mathbf{A} \succ 0$. Prove that $\mathbf{D} \succcurlyeq 0$ if and only if $c - \mathbf{b}^T\mathbf{A}^{-1}\mathbf{b} \geq 0$.
Following the answer here I have factorized the matrix $\mathbf{D}$ as:
$\begin{bmatrix} 1 & 0\\ b^TA^{-1} & 1 \end{bmatrix} \begin{bmatrix} A & 0 \\ 0 & c - b^TA^{-1}b^T \end{bmatrix} \begin{bmatrix} 1 & 0 \\ bA^{-1} & 1 \end{bmatrix}^T$
However, I'm not entirely sure how to complete the proof. I know that if we can show $\textbf{Tr}(\mathbf{D}) \geq 0$ and $\det (\mathbf{D}) \geq 0$, then we know that $\mathbf{D}$ is positive semidefinite, and it's easy to see that latter is true if and only if $c - \mathbf{b}^T\mathbf{A}^{-1}\mathbf{b} \geq 0$, but I'm not sure about the former. Maybe there's also a cleaner way to go about doing this?