Let $\mathbf M=\left[\begin{matrix}\bf A & \bf B\\\bf B^\top & \bf D \\\end{matrix}\right]$ be a symmetric n.n.d. matrix where $\bf A$ is square. Show that $\det\bf M\le\det A\det D$.
I guess this result is related to the standard inequality $\det \mathbf A\le\prod_{i=1}^na_{ii}$ for an n.n.d. matrix $\mathbf A=((a_{ij}))$ of order $n$.
It can be shown that $\det\mathbf M=\det\mathbf A\det(\mathbf D-\mathbf{B^\top A^{-1}B})$ when $\bf A$ is nonsingular.
Moreover, $\bf M$ will be n.n.d. iff $\bf A$ and $\mathbf D-\mathbf{B^\top A^{-1}B}$ are n.n.d., as is seen from this post.
Now, $\det(\mathbf D-\mathbf{B^\top A^{-1}B})=\det(\mathbf D -\mathbf C)$, say
$=\det \mathbf D\det(\mathbf I-\mathbf{D^{-1}C})$
$=(\det\mathbf D)\prod_i(1-d_{i})$ where $d_{i}$'s are the eigenvalues of $\mathbf{D^{-1}C}$.
If I could somehow conclude that the $d_i$'s are nonnegative (or possibly the matrix $\mathbf{D^{-1}C}$ is n.n.d.), then the product $\prod_i(1-d_{i})$ becomes bounded above by $1$ and my required inequality follows. Is this somewhat apparent? Also, it is seen that equality should hold in the required inequality iff $\mathbf B=\mathbf O$, which means that with my argument the $d_i$'s must be zero. Is this the correct way to proceed?