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.
 A: 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.
A: Here is a proof using Gram determinants. A matrix $(a_{ij})$ is positive definite if an only if there exists a system of linearly independent vectors $v_1$, $\ldots$, $v_n$ such that $a_{ij} = \langle v_i, v_j\rangle$. Now what we have to show is
$$G(v_1, \ldots, v_n) \le G(v_1, \ldots, v_m) \cdot G(v_{m+1}, \ldots, v_n)$$
Note that $v$,$v_1$, $\ldots$, $v_k$ for a system of vectors we have
$$d(v, \langle v_1, \ldots, v_k\rangle) ^2 = \frac{G(v, v_1, \ldots, v_k)}{G(v_1, \ldots, v_k)}$$
Using this equality for the distance from $w$ to the space generated by $w_1$, $\ldots$, $w_k$ we conclude:
If we enlarge a system of vectors from $K$ to $L$, then
$$\frac{G(w, K)}{G(K)} \ge \frac{G(w,L)}{G(L)}$$
Now, by induction ( telescoping product) if $I\supset I'$ and $J$ are system of vectors then
$$\frac{G(I)}{G(I')}\ge \frac{G(I\sqcup J)}{G(I'\sqcup J)}$$
In particular, if $I'$ is empty we get
$$G(I) \ge \frac{G(I\sqcup J)}{G(J)}$$
which is our inequality.
A: Here is a proof using Ky Fan maximum principle, up to a block unitary congruence $ \left(
\begin{array}{cc}
U&0
\\0 &V
\end{array}
\right) $ assume that $A=diag(d_1,\ldots,d_n)$ and $B=diag(d_{n+1},\ldots,d_{2n})$, further more assume without loss of generality that the variables are arranged in decreasing order ($d_1\ge d_2\ge \cdots \ge d_{2n}\ge 0$) then by Ky Fan maximum eigenvalue  principle $\sum_{i=1}^k\lambda_i(M)\ge \sum_{i=1}^k d_i$ for $1\le k \le 2n$ with equality when $k=2n$. This implies from majorization (as all variables are non negative) that $$\prod_{i=1}^{2n}d_i\ge \prod_{i=1}^{2n}\lambda_i(M).$$
