Suppose I have matrices $A, B, D \in \mathbb{R}^{n \times n}$, where $D$ is symmetric and positive semidefinite. Furthermore, let $\alpha_1 \in [0,1], \alpha_2 \in [0,2]$ be known scalars. Is there any condition I could put on $A$ and $B$ to ensure the following?
\begin{equation} \begin{bmatrix} \alpha_1 D & - \alpha_1 D \\ -\alpha_1 D & \alpha_2 D \end{bmatrix} - \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}^T \begin{bmatrix} D & D \\ D & D \end{bmatrix} \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix} \leq 0 \end{equation}
I tried to explore the Schur complement which works quite fine for the positive semidefinite counterparts, but for negative semidefiniteness, it seems to be quite hard.
You could write this in the form $P - Q^TDQ \leq 0$, where $$ P = \begin{bmatrix} \alpha_1 D & - \alpha_1 D \\ -\alpha_1 D & \alpha_2 D \end{bmatrix}, \quad Q =\begin{bmatrix} A & B \end{bmatrix}. $$ If $D$ is known to be invertible, then this is the Schur complement of $D^{-1}$ within the matrix $$ M = \begin{bmatrix} D^{-1} & Q\\ Q^T & P \end{bmatrix} = \begin{bmatrix} D^{-1} & A & B\\ A^T & \alpha_1 D & -\alpha_1 D\\ B^T & -\alpha_1 D & \alpha_2 D \end{bmatrix} $$ Thus, your condition is equivalent to $M$ having $n$ positive eigenvalues and $2n$ non-positive eigenvalues.
Note that the block-matrix $P$ can be written as the Kronecker product $$ P = \begin{bmatrix} \alpha_1 & -\alpha_1\\ -\alpha_1 & \alpha_2 \end{bmatrix}\otimes D. $$ As such, we can see that $P$ will have the maximal possible number of non-negative eigenvalues (excluding the case where $\alpha_1 = \alpha_2 = 0$) when $\alpha_1 \alpha_2 - \alpha_1^2 \leq 0 \implies \alpha_2 \leq \alpha_1$; the original condition cannot hold otherwise. In this case, and if $P$ is inveritble, $M$ will satisfy the desired condition if and only if the Schur complement of $P$ is negative definite.
Putting it all together, we can say that if $D$ is invertible and $\alpha_2 < \alpha_1$, then your condition is equivalent to $$ D^{-1} - QPQ^T \leq 0\implies\\ D^{-1} - \alpha_1(ADA^T) + \alpha_1(ADB^T + BDA^T) - \alpha_2 BDB^T \leq 0. $$