Let $A$ a square real matrix such that the largest singular value $\sigma\_\text{max}(A) = \sigma < 1$. I want to find a lower bound on $\langle (I + A)^{-1}x, x\rangle$ where $x$ is a vector of euclidean norm $1$: $\langle x, x\rangle=1$.


I empirically find that a seemingly tight lower bound is
$$
\langle (I + A)^{-1}x, x\rangle \geq \frac{1}{1+\sigma}
$$
which is reached for $A= \sigma I$. I cannot prove the above result.


Note that it is pretty straightforward to prove that $\sigma\_\text{min}((I + A)^{-1}) \geq \frac1{1+\sigma}$ but that does not suffice to conclude, since I do not assume that $A$ is symmetric.


PS: $\sigma\_\text{max}$ and $\sigma\_\text{min}$ are the largest and smallest singular values: $\sigma\_\text{max} = \sqrt{\lambda\_\text{max}(AA^T)}$ is the operator norm of $A$.



Lower bound $\langle (I + A)^{-1}x, x \rangle$ given that $\sigma_\text{max}(A) < 1$
The map $f(z)=(1+z)^{-1} - (1+\sigma)^{-1}$ maps the disk of radius $\sigma$ into the right half plane as a function of one complex variable.


Therefore, essentially by von Neumann's inequality, we get that $$\frac{f(A)+f(A)^\*}{2}=\mathrm{Re }f(A)\geq 0$$ since $\|A\|\leq \sigma.$ Assuming $A$ has real entries, this implies the claim as
$$\langle (1+A)^{-1}x,x\rangle = \langle \mathrm{Re} (1+A)^{-1}x,x\rangle.$$


To see the calculation with von Neumann's inequality more explicitly, let $\psi(z) = \frac{z-1}{z+1}.$ Note $\psi$ takes the right half plane to the disk. So, $\psi \circ f$ takes the disk of radius $\sigma$ into the unit disk.
Therefore, von Neumann's inequality states that
$\|\psi \circ f(A)\|\leq \sup\_{z\in \sigma\mathbb{D}} |\psi\circ f(z)| \leq 1.$
Note that $$1-(\psi \circ f(A))^\*(\psi \circ f(A)) \geq 0.$$ (Here by $T\geq 0$ we mean that $T$ is positive semi-definite.)
Writing out what that means
$$1-(f(A)^\*+1)^{-1}(f(A)-1)^\*(f(A)-1)(f(A)+1)^{-1} \geq 0.$$
So,
$$(f(A)^\*+1)(f(A)+1)-(f(A)^\*-1)(f(A)-1)=2(f(A)+f(A)^\*)\geq 0.$$


Results of the above form (positivity of noncommutative rational functions) always have to have "algebraic proofs," many of which can be done algorithmically. See, e. g.,
[Helton, Klep, and McCullough - The convex Positivstellensatz in a free algebra](https://www.sciencedirect.com/science/article/pii/S0001870812001892) and
[Pascoe - Positivstellensätze for noncommutative rational expressions](https://www.ams.org/journals/proc/2018-146-03/S0002-9939-2017-13773-3/home.html).


