What is $\frac{\iota'(I+A)^{-2}\iota}{\iota'(I+A)^{-1}\iota}$ for A p.d.? Le $A\in\mathbb{R}^{N\times N}$ positive definite and $I$ the identity matrix. What is $\frac{\iota'(I+A)^{-2}\iota}{\iota'(I+A)^{-1}\iota}$ where $\iota$ is a vector of ones?

Let the eigenvalues of $A$ be $d_1\ge d_2\ge \dots d_N>0$. Then your ratio can be as small as $1/(1+d_1)$ and as large as $1/(1+d_N)$, or anything in between, depending on how $\iota$ lines up with the eigenvectors of $A$.

In particular, write $A=O^t \Delta O$ where $\Delta$ is diagonal and $O$ orthogonal. Then $$\iota^t(I+A)^{-1}\iota = w^t (I+\Delta)^{-1}w = \sum w_i^2/(1+d_i),$$ where $w=O\iota$, etc. So your ratio is $$R=\frac{\sum_i w_i^2/(1+d_i)^2}{\sum_i w_i^2 / (1+d_i)}.$$ The weights $w_i$ are constrained by $\|w\|^2=\|\iota\|^2=N$, and by putting all the mass at $i=1$ one obtains $R=1/(1+d_1)$ and by putting all the mass at $i=N$ one obtains $R=1/(1+d_N)$.