I am interested in the following function, $$ f(P, u) := \mathrm{tr}\Big((P + I)^{-1}(P + d uu^T)(P + I)^{-1}\Big). $$ Above, $P$ is a symmetric positive definite real matrix and $u$ is a vector with unit norm. Here $d$ denotes the dimension. It is a positive integer.
I can use the bound $uu^T \leq I$ in the positive definite ordering and I obtain $$ f(P, u) \leq \mathrm{tr}\Big((P + I)^{-1}(P + d I)(P + I)^{-1}\Big), $$ however I do not believe this bound can be attained for a unit vector $u$ (simply because $uu^T$ is rank 1).
Is there a better characterization of the following maximal quantity? $$ \sup_{u : \|u\| = 1} f(P, u) $$
Note that we have $$ \begin{align} \operatorname{tr}[(P + I)^{-1}(P &+ duu^T)(P + I)^{-1}] \\&= \operatorname{tr}[(P + I)^{-1}P(P + I)^{-1} + d(P + I)^{-1}uu^T(P + I)^{-1}] \\ & = \operatorname{tr}[(P + I)^{-1}P(P + I)^{-1}] + d\operatorname{tr}[(P + I)^{-1}uu^T(P + I)^{-1}] \\ & = \operatorname{tr}[(P + I)^{-1}(P + I)^{-1}P] + d\operatorname{tr}[u^T(P + I)^{-1}(P + I)^{-1}u] \\ & = \operatorname{tr}[(P + I)^{-2}P] + d\,u^T(P + I)^{-2}u \\&= \operatorname{tr}[(P + I)^{-2}P] + d\cdot \|(P + I)^{-1}u\|^2. \end{align} $$ So, if $d > 0$, it's clear that $$ \sup_{u:\|u\|=1} f(P,u) = \operatorname{tr}[(P + I)^{-2}P] + d \sup_{u:\|u\|=1} \|(P + I)^{-1}u\|^2. $$ Because $P$ is symmetric an positive definite, this turns out to be easy to compute in terms of the eigenvalues (which equal the singular values) of $P$. We have $$ \sup_{u:\|u\|=1} \|(P + I)^{-1}u\|^2 = \frac{1}{(1 + \lambda_{\min}(P))^2}, $$ leading to the overall maximum $$ \sup_{u:\|u\|=1} f(P,u) = \operatorname{tr}[(P + I)^{-2}P] + \frac{d}{(1 + \lambda_{\min}(P))^2}. $$ If you prefer, we can rewrite $\lambda_\min(P) = \|P^{-1}\|^{-1}$, where $\|\cdot\|$ denotes the spectral norm.