Why does $(\det A)(\det C) \ge {\left| {\det B} \right|^2}$? Let $H = \left[ {\begin{array}{*{20}{c}}
   A & B  \\
   {{B^*}} & C  \\
\end{array}} \right]$ be positive semidefinite and $A,C\in M_p$.
Why does $(\det A)(\det C) \ge {\left| {\det B} \right|^2}$?
 A: Hint:  Assume that $H$ is positive-definite, and use continuity to cover the case where $H$ is positive-semidefinite.  If $H$ is positive-definite, then show that $A$ and $C$ also are, and that $A-BC^{-1}B^*$ is positive-definite.  Finally, prove that, if $X$ and $Y$ are positive-semidefinite matrices such that $X-Y$ is positive-semidefinite, then $\det(X)\geq\det(Y)$.
Disclaimer: This hint works only for complex positive-semidefinite matrices.  I haven't thought about the case where the ground field is the reals.
Further Hint:  Let $V$ be the matrix $\begin{bmatrix}I&-BC^{-1}\\0&I\end{bmatrix}$.  Then, $VHV^*=\begin{bmatrix}A-BC^{-1}B^*&0\\0&C\end{bmatrix}$.
A: First you show that if $A,C $ are invertible, 
$$
\begin{bmatrix}
A&B\\ B^*&C
\end{bmatrix}
=
\begin{bmatrix}
A^{1/2}&0\\ 0&C^{1/2}
\end{bmatrix}  
\begin{bmatrix}
I&A^{-1/2}BC^{-1/2}\\ C^{-1/2}B^*A^{-1/2}&I
\end{bmatrix} 
\begin{bmatrix}
A^{1/2}&0\\ 0&C^{1/2}
\end{bmatrix},
$$
so we may assume
$$
\begin{bmatrix}
I&A^{-1/2}BC^{-1/2}\\ C^{-1/2}B^*A^{-1/2}&I
\end{bmatrix}   
\geq0.
$$
If $G=A^{-1/2}BC^{-1/2} $, we have
\begin{align}
0&\leq \left\langle\begin{bmatrix}I&G\\ G^*&I\end{bmatrix}\begin{bmatrix}x\\ y\end {bmatrix},\begin{bmatrix}-Gx\\ x\end {bmatrix}\right\rangle=\|Gx\|^2+\|x\|^2-2\langle Gx,Gx\rangle \\ \ \\
&=\|x\|^2-\|Gx\|^2=\langle x,x\rangle-\langle G^*Gx,x\rangle.
\end{align}
This implies that all eigenvalues of $G^*G $ are in $(0,1] $. Thus, so is their product: $\det G^*G\leq1$. Then
$$
\frac {|\det B|^2}{(\det A)(\det C)}=\det (C^{-1/2}B^*A^{-1}BC^{-1/2})=\det (G^*G)\leq1.
$$
Finally, if we do not assume invertibility of  $A $ and $C $, we do the above for $A+\varepsilon  I $ and $C+\varepsilon I $ for arbitrary  $\varepsilon >0$.
