Determinant of PSD matrix and PSD submatrix inequality I'm reading this paper and in the appendix I see the following statement:
For $A \in R^{m\times m}, B \in R^{n\times m}, C \in R^{n\times n}$,
if $D = \begin{bmatrix}A & B\\B^T & C\end{bmatrix}$ is positive semi-definite then,
$det(D) \leq det(A)det(C)$
This is given without proof as a property of psd matrices. This doesn't seem axiomatic to me and it's not obvious.  Can you point to a reference or give a proof of this? I suspect it's pretty simple, but I'm missing it.  I've never formally studied linear algebra so it might just be a gap in my education.
Some things I notice:
$A$ and $C$ are principal submatrices of $D$.
I know a determinant of a $2\times 2$ matrix is $a_{1,1}a_{2,2} - a_{1,2}a_{2,1}.$ 
Because $D$ is psd and has larger dimensions than $A$ or $C$, it seems like the second term is subtracting more than the second term for $A$ or $C$ would.  But that statement is pretty imprecise and doesn't convince me that it's true.

Lemma: If $A$ and $B$ are symmetric positive-definite, then $\det(A+B) \geq \det(A)$.
This follows from Sylvester's determinant theorem: if $L$ is a Cholesky factor of $B$,
$$\det(A+B) = \det(A)\det(I + L^TA^{-1}L) \geq \det(A)$$
since $L^TA^{-1}L$ is symmetric positive-definite, and adding $I$ shifts the spectrum by one.

Now for your original problem, if $D$ is singular the statement is obvious. If it's strictly positive-definite, write
$$D=\left[\begin{array}{cc}A & 0\\B^T & I\end{array}\right]\left[\begin{array}{cc}I & A^{-1}B\\0 & C - B^TA^{-1}B\end{array}\right].$$
The matrix $C-B^T A^{-1}B$ is symmetric, and also positive-definite, since if $v$ is an eigenvector with negative eigenvalue $\lambda$, we have
$$(-A^{-1}Bv\quad v)^T\,D\,(-A^{-1}Bv\quad v) = (-A^{-1}Bv\quad v)^T(0 \quad \lambda v) = \lambda\|v\|^2<0.$$
Thus by the lemma, $\det(C) \geq \det(C-B^TA^{-1}B)$, and your statement follows.
