Let $A$ be a SPD matrix, let $D$ be the diagonal of $A$, and let $G$ be an arbitrary matrix of the proper size (not necessarily full rank). Do we have the following inequality$$\det(I+G^\top A^{-1}G) \leq \det(I+G^\top D^{-1}G) $$ I feel like this is not true. We can surely prove that $\det(A^{-1})\leq \det(D^{-1})$, but I don't know if this the inequality hold in general.
Also, if this is not true, what condition do we have to impose to $G, A, D$ for this inequality to hold?
The A produced by @user1551, namely $A=\begin{pmatrix} 2 &2\\ 1 & 2\\ \end{pmatrix}$ combined with $G = I$, produces a counterexample:
$\det(I+G^T A^{-1}G) = 8/3 > 9/4 = \det(I+G^T D^{-1}G)$
$\det(I+G^T A^{-1}G) \le \det(I+G^T D^{-1}G)$ is trivially true when $G =$ zero matrix and/or $A$ is a diagonal matrix.