if $M=\left( \begin{matrix} A &B \\B^{T} &C \end{matrix} \right)$ is a positive definite matrix,Prove that $|M|\leq|A||C|$ 
Let $A,B,C$ be n-order matrices. If $$M=\left(
\begin{matrix}
A &B
\\B^{T} &C
\end{matrix}
\right)$$ is a positive definite matrix, prove that $|M|\leq|A||C|$.

My attempt: Since A and C is necessarily also positive definite, there is some invertible $P$ and $Q$ such that $P^{T}AP=E_n$ and $Q^{T}CQ=E_n$, hence we have $$\left(
\begin{array}{cc}
P^{T} &0
\\0 &Q^{T}
\end{array}
\right)
\left(
\begin{array}{cc}
A &B
\\B^{T} &C
\end{array}
\right)
\left(
\begin{array}{cc}
P &0
\\0 &Q
\end{array}
\right)=\left(
\begin{array}{cc}
E_n &P^TBQ
\\Q^TB^TP &E_n
\end{array}
\right)$$ so taking det both sides we have $|M||A||C|\leq \det(RHS)$, but I'm stuck here. Does anyone know how to prove it? Thank you.

To finish the proof via congruence:
$P^TAP = I_n$ and  $Q^TCQ = I_n$
$\implies \det\big(P^TP\big) = \det\big(A\big)^{-1}$and $\det\big(Q^TQ\big) = \det\big(C\big)^{-1}$
$\mathbf 0\prec \left[
\begin{array}{cc}
P^{T} &\mathbf 0
\\\mathbf 0 &Q^{T}
\end{array}
\right]
\left[
\begin{array}{cc}
A &B
\\B^{T} &C
\end{array}
\right]
\left[
\begin{array}{cc}
P &\mathbf 0
\\\mathbf 0 &Q
\end{array}
\right]=\left[
\begin{array}{cc}
I_n &*
\\* &I_n
\end{array}
\right]$
taking determinants:
$\det\left(\left[
\begin{array}{cc}
P^{T} &\mathbf 0
\\\mathbf 0 &Q^{T}
\end{array}
\right]\right)
\det\left(\left[
\begin{array}{cc}
A &B
\\B^{T} &C
\end{array}
\right]\right)
\det\left(\left[
\begin{array}{cc}
P &\mathbf 0
\\\mathbf 0 &Q
\end{array}
\right]\right)$
$=\det\left(\left[
\begin{array}{cc}
P^{T}P &\mathbf 0
\\\mathbf 0 &Q^{T}Q
\end{array}
\right]\right)
\cdot\det\big(M\big)$
$=\det\big(P^TP\big)\cdot \det\big(Q^TQ\big)\cdot\det\big(M\big)$
$=\det\big(A\big)^{-1}\cdot \det\big(C\big)^{-1}\cdot\det\big(M\big)$
$=\det\left(\left[
\begin{array}{cc}
I_n &*
\\* &I_n
\end{array}
\right)\right]$
$\leq 1$
by Hadamard's Determinant Inequality.
So $\det\big(A\big)^{-1}\cdot \det\big(C\big)^{-1}\cdot\det\big(M\big)\leq 1$ and re-scaling each side by $\det\big(A\big)\cdot \det\big(C\big)$ gives the result.
note:
There is no need for $A$ and $C$ to be the same size.  This proof runs verbatim when $M\succ \mathbf 0$ and $A$ and $C$ are square.  With this view: a special case of this inequality occurs when $C$ is $1\times 1$, which was this question:
Inequality for a determinant
Via induction, said special case proves Hadamard's Determinant Inequality which is at the core of this problem.
