Define $$ \mathcal{K}(A,B)=\begin{bmatrix}A&I\\I&B\end{bmatrix} $$ where $I$ is the identity matrix with proper dimensions,
I have this optimization problem:
$$ \begin{aligned} \min_{X,S,W,V}&\text{Trace}(VX+WS) \\ \text{subject to }& \mathcal{K}(X,S)\geq 0\\ & \mathcal{K}(V,W)\geq 0\\ &V=V^T\in\mathbb{R}^{n\times n}\\ &X=X^T\in\mathbb{R}^{n\times n}\\ &W=W^T\in\mathbb{R}^{n\times n}\\ &S=S^T\in\mathbb{R}^{n\times n}\\ \end{aligned} $$ Seems that the analytic minimum of the problem is $2n$, yet I failed to show that. Any help is appreciated! $\geq$ means "postive semidefinite".
Minimizing first over $X$, we know from an answer to your previous question that the positive semidefinite constraints imply $X \geq S^{-1} > 0$ and so we find that after minimizing over $X$ the objective function becomes $$ \mathrm{Tr}[VS^{-1} + W S]. $$ Using the same logic, after minimizing over $V$ we arrive at $$ \mathrm{Tr}[W^{-1}S^{-1} + W S] = \mathrm{Tr}[W^{-1/2}S^{-1} W^{-1/2} + W^{1/2} S W^{1/2}]. $$ Writing $Y = W^{1/2} S W^{1/2}$ the problem is then equivalent to $$ \begin{aligned} &\min \quad\mathrm{Tr}[Y + Y^{-1}]\\ &\mathrm{s.t.}\qquad Y>0 \end{aligned} $$ Indeed for any $Y>0$, $(X,S,W,V) = (Y^{-1},Y, I, I)$ is a feasible point of the original problem such that $Y = W^{1/2} S W^{1/2} > 0$. Conversely $W^{1/2} S W^{1/2} > 0$ as $S > 0$ by the constraints of the original problem. Thus we need only solve this final problem.
Let $\{\lambda_i\}_i$ be the eigenvalues of $Y$. Then by the spectral theorem the problem is equivalent to $$ \begin{aligned} \min& \quad\sum_i \lambda_i + \frac{1}{\lambda_i} \\ \mathrm{s.t.}& \qquad \lambda_i > 0 \qquad \text{for all i} \end{aligned} $$
We can minimize each term in the sum separately and it is not difficult to show that it is minimized at $\lambda_i = 1$. Thus we have the objective value is $\sum_{i=1}^n 2 = 2n$.
Note that this can be achieved in the original optimization by choosing the feasible point $(I,I,I,I)$.