This questions concerns the practical implementation of schur's complement and Young's inequality.
Consider the following
\begin{align} \begin{pmatrix} \begin{pmatrix} \mathbb{A}_{\mathbb{Z}} & \mathcal{O}\\ \mathcal{O} & \mathbb{A}_{\mathbb{P}} \end{pmatrix} + \begin{pmatrix} \mathbb{Z} & \mathcal{O}\\ \mathcal{O} & \mathcal{I}_{n} \end{pmatrix} \begin{pmatrix} \mathbb{Z} & \mathcal{O}\\ \mathcal{O} & \mathcal{I}_{n} \end{pmatrix} & \begin{pmatrix} D_{11} & D_{12}\\ D_{21} & D_{22} \end{pmatrix}\\ \begin{pmatrix} D_{11} & D_{12}\\ D_{21} & D_{22} \end{pmatrix}^T & -\mu \mathbb{I}_{q} \end{pmatrix} + \underbrace{\begin{pmatrix} \overline{B} \overline{K}\\ \mathcal{O}\\ \mathcal{O} \end{pmatrix}}_{V^T} \underbrace{\begin{pmatrix} \mathcal{O} & \tilde{B} & \mathcal{O} \end{pmatrix}}_U + \begin{pmatrix} \mathcal{O}\\ \tilde{B}^T\\ \mathcal{O} \end{pmatrix} \begin{pmatrix} \left(\overline{B} \overline{K} \right)^T & \mathcal{O} & \mathcal{O} \end{pmatrix} \leq 0 \end{align}
Now to avoid any bilinear coupling between terms, the young's inequality is used, as given by
\begin{equation} X^T \hspace{1mm} Y + Y^T \hspace{1mm} X \leq \dfrac{1}{2} \hspace{1mm} \left( X + S \hspace{1mm} Y \right)^T \hspace{1mm} S^{-1} \hspace{1mm} \left( X + S \hspace{1mm} Y \right) \end{equation}
where $S$ is a symmetric positive definite matrix. To apply the inequality, we use $S = \dfrac{1}{2} \mathbb{I}_n$ for the term $\begin{pmatrix} \mathbb{Z} & \mathcal{O}\\ \mathcal{O} & \mathcal{I}_{n} \end{pmatrix} \begin{pmatrix} \mathbb{Z} & \mathcal{O}\\ \mathcal{O} & \mathcal{I}_{n} \end{pmatrix}$ and $S = \epsilon \mathbb{Z}$ for the term $V^TU+U^TV$. Now, the intent is to use the young's inequality and convert the above algebraic equation in LMI using Schur's complement. I am unsure of the proper way and cannot reduce the equation to the requisite LMI. Given: $\mathbb{Z}$ is a $n \times n$ symmetric positive definite matrix.
Can someone walk me through the subsequent steps?
The only term that is not linear in the unknown variables is the quadratic term in $\mathbb{Z}$, so it might be easier to replace that with another variable and add another inequality
$$ \mathbb{Z}^2 \leq \mathbb{X}, \tag{1} $$
such that the initial in equality can be written as
\begin{align} \begin{pmatrix} \begin{pmatrix} \mathbb{A}_{\mathbb{Z}} & \mathcal{O}\\ \mathcal{O} & \mathbb{A}_{\mathbb{P}} \end{pmatrix} + \begin{pmatrix} \mathbb{X} & \mathcal{O}\\ \mathcal{O} & \mathcal{I}_{n}^2 \end{pmatrix} & \begin{pmatrix} D_{11} & D_{12}\\ D_{21} & D_{22} \end{pmatrix}\\ \begin{pmatrix} D_{11} & D_{12}\\ D_{21} & D_{22} \end{pmatrix}^\top & -\mu\,\mathbb{I}_{q} \end{pmatrix} + \begin{pmatrix} \overline{B} \overline{K}\\ \mathcal{O}\\ \mathcal{O} \end{pmatrix} \begin{pmatrix} \mathcal{O} & \tilde{B} & \mathcal{O} \end{pmatrix} + \begin{pmatrix} \mathcal{O}\\ \tilde{B}^\top\\ \mathcal{O} \end{pmatrix} \begin{pmatrix} \left(\overline{B} \overline{K} \right)^\top & \mathcal{O} & \mathcal{O} \end{pmatrix} \leq 0, \tag{2} \end{align}
which is linear in all variables. One can then use Schur's complement on $(1)$ which allows us to rewrite it also as an LMI
$$ \begin{pmatrix} I & \mathbb{Z} \\ \mathbb{Z} & \mathbb{X} \end{pmatrix} \geq 0, \tag{3} $$
with $I$ an $n$ by $n$ identity matrix, $\mathbb{Z} = \mathbb{Z}^\top > 0$ and $\mathbb{X} = \mathbb{X}^\top > 0$.
So the combination of $(2)$ and $(3)$ should be equivalent to your initial inequality and are all LMI's.