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 $X$.

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.
