Let $a, b > 0$ and let $X, Y$ be symmetric positive semidefinite matrices. I want to show that $$\|aX(aX+bY+abI)^{-1}\|_{F}^{2} \leqq \|X(X+bI)^{-1}\|_{F}^{2}$$
My thought:
$$\|aX(aX+bY+abI)^{-1}\|_{F}^{2} \leqq \|X(X+bI)^{-1}\|_{F}^{2} \|a(X+bI)(a(X+bI)+bY)^{-1}\|_{2}^{2}$$
But,
$$\|a(X+bI)(a(X+bI)+bY)^{-1}\|_{2}^{2}$$
doesn't have any upper bound. Do you have any ideas? Or is it the case that this hypothesis does not hold?
This is not true. Let $Z=\frac baY$. Your inequality is equivalent to $$ \|X(X+Z+bI)^{-1}\|_F\le\|X(X+bI)^{-1}\|_F $$ for all PSD matrices $X$ and $Z$. If this were true, then by passing $b$ to the limit zero, we would obtain, in particular, that $\|X(X+Z)^{-1}\|_F\le\|X(X)^{-1}\|_F=\|I\|_F$ whenever $X,Z\succeq0$ and $(X+Z)^{-1}$ exists. That is, the norm of $X(X+Z)^{-1}$ is bounded above. Yet this is demonstrably false, as shown in a counterexample I gave yesterday.