I have the following matrix inequality that I need to express as an LMI
$ A^T Q^{-1} A - Q + \sum_i A_i^T Q^{-1} A_i < 0$
($Q > 0$, $A = XQ+YB$, $A_i = X_i Q+Y_i B$)
which I rewrote to
$ A^T Q^{-1} A - Q < - \sum_i A_i^T Q^{-1} A_i$
Now I want to use Schur's complement and express this inequality as
$ \begin{bmatrix} -Q & A^T \\ A & -Q \end{bmatrix} < \sum_i \begin{bmatrix} 0 & A_i^T \\ A_i & Q \end{bmatrix} $
and therefore
$ \begin{bmatrix} -Q & A^T \\ A & -Q \end{bmatrix} - \sum_i \begin{bmatrix} 0 & A_i^T \\ A_i & Q \end{bmatrix} < 0 $
which is a form my solver, solving for Q and B, should be able to handle.
Can I use the Schur complement in this way and are these two inequalities equivalent?
You can't use Schur's complement in that way. Namely when you have the matrix inequalities
$$ A - B^\top C^{-1} B \succ 0, \quad C \succ 0, $$
then you can combine this into the following one matrix inequality
$$ M = \begin{bmatrix} A & B^\top \\ B & C \end{bmatrix} \succ 0. $$
This is because $M$ can be decomposed as follows
$$ M = \underbrace{ \begin{bmatrix} I & B^\top C^{-1} \\ 0 & I \end{bmatrix} }_{T^\top} \begin{bmatrix} A - B^\top C^{-1} B & 0 \\ 0 & C \end{bmatrix} \underbrace{ \begin{bmatrix} I & 0 \\ C^{-1} B & I \end{bmatrix} }_{T}. $$
Namely $M \succ 0$ means that $x^\top M\,x > 0\ \forall\, x\neq0$. So when using $x=T\,y$ one gets $x^\top M\,x = y^\top T^\top M\,T\,y$. It holds that $x\neq0$ iff $y\neq0$, since $T$ is full rank (lower triangular with ones on the diagonal), thus it also has to hold that $T^\top M\,T \succ 0$.
But remember that this equivalent matrix inequality formulation only works because the matrix $T$ transforms $M$ into a block diagonal matrix with the blocks the original matrix inequalities. So your proposition only works if there exists a transformation $T$ such that
$$ T^\top \left( \begin{bmatrix} Q & -A^\top \\ -A & Q \end{bmatrix} + \sum_i \begin{bmatrix} 0 & A_i^\top \\ A_i & Q \end{bmatrix} \right) T $$
gives back the original matrix inequalities as blocks on the diagonal.