Inequality of Positive-definite matrix. In this question matrix $A$ is positive-definite if and only if $\forall x\ne0 :x^TAx>0$. ($A$ is not necessarily symmetric)
Let $D$ be a positive-definite matrix such that it has block form:
$$D=\left( \begin{array}{cc}
A & C  \\
C^T & B  \end{array} \right)$$
How can we prove that $\det D\leq\det A\det B$? 
EDIT1: From the perspective of the first answer I want to sum up something.


*

*It's true that 


$$D=\left( \begin{array}{cc}
A & C  \\
C^T & B  \end{array} \right)=\left( \begin{array}{cc}
I & 0  \\
C^TA^{-1} & I  \end{array} \right)\left( \begin{array}{cc}
A & C  \\
0 & B-C^TA^{-1}C  \end{array} \right)$$
so $\det D=\det A\det (B-C^TA^{-1}C)$. 
It's also true that $\det A>0$ because every real eigenvalues of $A$ is greater than $0$ and all complex eigenvalues exist in pairs.
Now we need to prove that $$\det B>\det(B-C^TA^{-1}C)$$


*

*The usual way when $A$ and $B$ are symmetric haven't worked yet because: (1) we don't know for sure if $(B-C^TA^{-1}C)$ is definite positive and (2) we don't know for sure if $\det(M+N)>\det(N)$ if $M$ and $N$ are definite positive.


I also want to point out that definite positive matrices (in this question) can have complex eigenvalues. 
It would be great if you answer with details, not with references since almost every references consider positive-definite matrices to be symmetric.
 A: What you want to prove is not necessarily true. Here is a random counterexample:
$$
D=\left[\begin{array}{c|c}A&C\\ \hline C^T&B\end{array}\right]
=\left[\begin{array}{rr|rr}
2&-3&1&0\\
2&2&0&-1\\
\hline
1&0&2&1\\
0&-1&-2&2
\end{array}\right].
$$
According to your definition, a matrix is "positive definite" if and only if its symmetric part is positive definite in the conventional sense. Now, WolframAlpha reckons that the eigenvalues of
$$
D+D^T=\left[\begin{array}{rr|rr}
4&-1&2&0\\
-1&4&0&-2\\
\hline
2&0&4&-1\\
0&-2&-1&4
\end{array}\right]
$$
are $4-\sqrt{5}$ and $4+\sqrt{5}$ (each of multiplicity 2). Hence $D$ is "positive definite". WolframAlpha also reckons that $\det(D)=61$. However,
$$\det(A)\det(B)=10\times6=60<\det(D).$$
A: You need two facts to prove the result. One is about the determinant of the block matrices.
$$\det D = \det A \det (B-C^\top A^{-1}C)$$
The other one is called Schur's complement.
See page 66  here.
$$D > 0 \leftrightarrow A > 0, B - C^\top A^{-1}C >0 $$
Note that $C^\top A^{-1}C \geq 0$ due to $A > 0$. This gives $B > 0$. From this you get 
$$\det (B-C^\top A^{-1}C) \leq \det B$$
Hence, $$\det D \leq \det A \det B$$
Edit: The determinant inequality follows from this
Edit #2 (Proof of $D > 0 \rightarrow A > 0 $): We assume $A$ to be square, $A \in \mathbb{R}^{n \times n}$. Note that when $A$ is square, it follows from the structure of $D$ that $B$ is square as well, $B \in \mathbb{R}^{m \times m}$. Take $x \in  \mathbb{R}^{n}$.
$$\begin{pmatrix} x^{\top} & 0 \end{pmatrix}D\begin{pmatrix} x \\ 0 \end{pmatrix} = x^{\top}Ax$$
Since $D$ is positive definite ($y^{\top}Dy >0 \quad \forall y \in \mathbb{R}^{(n+m) \times (n+m)}$), $x^{\top}Ax > 0 \quad \forall x \in \mathbb{R}^{n \times n}$. Hence, $A$ is positive definite.
For the implication, $D > 0 \rightarrow B - C^\top A^{-1}C >0 $ I haven't been able to come up with a proof that doesn't require symmetricity assumption.
