Does the Schur complement preserve the partial order? Let 
$$\begin{bmatrix}
A_{1} &B_1  \\ B_1'  &C_1
\end{bmatrix} \quad \text{and} \quad \begin{bmatrix}
A_2 &B_2 \\ B_2'  &C_2
\end{bmatrix}$$
be symmetric positive definite and conformably partitioned matrices. If 
$$\begin{bmatrix}
A_{1} &B_1  \\ B_1'  &C_1
\end{bmatrix}-\begin{bmatrix}
A_2 &B_2 \\ B_2'  &C_2
\end{bmatrix}$$
is positive semidefinite, is it true 
$$(A_1-B_1C^{-1}_1B_1')-(A_2-B_2C^{-1}_2B_2')$$ also positive semidefinite? Here, $X'$ means the transpose of $X$.
 A: Yes, it does. The assumption $$\begin{bmatrix}
A_{1} &B_1  \\ B_1^T  &C_1
\end{bmatrix}-\begin{bmatrix}
A_2 &B_2 \\ B_2^T  &C_2
\end{bmatrix} \geq 0$$ implies that for any vector $\begin{pmatrix} x & y \end{pmatrix}$,
$$ \begin{pmatrix} x^T & y^T \end{pmatrix} \begin{bmatrix}
A_{1} &B_1  \\ B_1^T  &C_1
\end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \geq \begin{pmatrix} x^T & y^T \end{pmatrix} \begin{bmatrix}
A_2 &B_2 \\ B_2^T  &C_2
\end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix}~~~~~(*)$$ But for any partitioned matrix, 
$$\begin{pmatrix} x^T & y^T \end{pmatrix} \begin{bmatrix}
A &B  \\ B^T  &C
\end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = (x + A^{-1} B y)^T A (x + A^{-1} B y) + y^T(C-B^T A^{-1} B)y.$$ Moreover, if the partitioned matrix on the left-hand side is positive definite, then each of the two terms on the right=hand side is positive. Thus picking arbitrary $y$ and $x = -A_1^{-1}B_1y$ in (*) gives
$$y^T(C_1-B_1^T A_1^{-1} B_1)y \geq \mbox{ something positive} + y^T(C_2-B_2^T A_2^{-1} B_2)y,$$ which implies 
$$y^T(C_1-B_1^T A_1^{-1} B_1)y \geq  y^T(C_2-B_2^T A_2^{-1} B_2)y,$$
which implies the conclusion you want.
A: In control theory, this kind of inequalities is ubiquitous and handled via going back and forth using Schur complements. For completeness, here is the non-strict version of Schur complement formula, it is an overkill but the question is a particular special case, so here it goes:
Formula: Let $Q,R$ be symmetric matrices. Then following are equivalent:


*

*$$
\begin{pmatrix}
Q &S\\
S^T &R
\end{pmatrix} \succeq 0
$$

*$$\begin{align}
R &\succeq 0\\
Q -SR^\dagger S^T &\succeq 0\\
S(I-RR^\dagger) &= 0
\end{align}
$$
where $R^\dagger$ is the Pseudoinverse of $R$.

Now if renaming your hypothesis involving the matrix difference as 
$$ M_1 - M_2 := \begin{pmatrix}
A_1 &B_1\\
B_1^T &C_1
\end{pmatrix} - \begin{pmatrix}
A_2 &B_2\\
B_2^T &C_2
\end{pmatrix}\succeq 0$$
we can reformulate the hypothesis as the following via the second item of the formula ($Q=M_1,S=I,R=M_2^{-1}$): 
$$\begin{align}
M_2^{-1} &\succeq 0 \quad \text{by definition}\\
M_1 - M_2 &\succeq 0\\
I (I-M_2M_2^{-1}) &= 0
\end{align}
$$
hence we have
$$
\begin{pmatrix}
M_1 &I\\I&M_2^{-1}
\end{pmatrix}\succeq 0
$$
Also, from the inverse of a matrix formula, we have
$$
M_2^{-1} = \begin{pmatrix} 
(A_2 - B_2C_2^{-1}B_2^T)^{-1} &\star\\
\star &\star
\end{pmatrix}
$$
As user1551 showed, you can bring the matrix $M_1$ in the form of the following by a congruence transformation and some reshuffling:
$$\begin{pmatrix}
M_1 &I\\I&M_2^{-1}
\end{pmatrix} \leadsto \left(
\begin{array}{cc|cc}
(A_1 - B_1C_1^{-1}B_1^T) &I &0 &0\\
I &(A_2 - B_2C_2^{-1}B_2^T)^{-1} &0 &\star\\ \hline
0 &0 &C_1 &I\\
0&\star&I&\star
\end{array}\right)
\succeq 0
$$
The $(1,1)$ block matrix has the desired result if we apply the nonstrict Schur complement formula again but in the reverse direction. 
Last minute addition: Now that I look at it, it is not as good as I thought it to be initially but I did not want to waste the whole thing so I hope it helps an $\epsilon$.
A: For a general block matrix $X=\begin{pmatrix}A&B\\C&D\end{pmatrix}$, the Schur complement $S$ to the block $D$ satisfies
$$
\begin{pmatrix}A&B\\C&D\end{pmatrix}
=\begin{pmatrix}I&BD^{-1}\\&I\end{pmatrix}
\begin{pmatrix}S\\&D\end{pmatrix}
\begin{pmatrix}I\\D^{-1}C&I\end{pmatrix}.
$$
So, when $X$ is Hermitian,
$$
\begin{pmatrix}A&B\\B^\ast&D\end{pmatrix}
=\begin{pmatrix}I&Y^\ast\\&I\end{pmatrix}
\begin{pmatrix}S\\&D\end{pmatrix}
\begin{pmatrix}I\\Y&I\end{pmatrix}\ \textrm{ for some } Y.
$$
Hence
$$
\begin{eqnarray}
&&\begin{pmatrix}A_1&B_1\\B_1^\ast&D_1\end{pmatrix}
\ge\begin{pmatrix}A_2&B_2\\B_2^\ast&D_2\end{pmatrix}
\\
&\Rightarrow&
\begin{pmatrix}S_1\\&D_1\end{pmatrix}
\ge
\begin{pmatrix}I&Z^\ast\\&I\end{pmatrix}
\begin{pmatrix}S_2\\&D_2\end{pmatrix}
\begin{pmatrix}I\\Z&I\end{pmatrix}\ \textrm{ for some } Z\\
&\Rightarrow&
(x^\ast,0)\begin{pmatrix}S_1\\&D_1\end{pmatrix}\begin{pmatrix}x\\0\end{pmatrix}
\ge
(x^\ast,\ x^\ast Z^\ast)
\begin{pmatrix}S_2\\&D_2\end{pmatrix}
\begin{pmatrix}x\\Zx\end{pmatrix},\ \forall x\\
&\Rightarrow&
x^\ast S_1 x
\ \ge\ x^\ast S_2 x + (Zx)^\ast D_2 (Zx)
\ \ge\ x^\ast S_2 x,\ \forall x\\
&\Rightarrow& S_1\ge S_2.
\end{eqnarray}
$$
Edit: In hindsight, this is essentially identical to alex's proof.
