I know that eigenvalue estimates involving products of matrices are in general tricky, but probably this question has some hope:


Let $A$ and $B$ be two real symmetric positive semi-definite $n\times n$ matrices (with $A+B$ positive definite if needed). Let $\lambda(X)$ denote any eigenvalue of $X$.


I am fairly sure that
$$
|1-\lambda\big( (I+A+B+AB)^{-1}(A+B)\big)| < 1
$$
but I would like ask:


* Do you know a slick proof of this?
* Do you know finer estimates (using,e.g., all eigenvalues/vectors of $A$, $B$ and $A+B$) or techniques that could be helpful?



Estimates of eigenvalues
Yes, and more may be said.


Assume that $(I+A+B+AB)^{-1}(A+B)z=\lambda z$, then $(I+c(A+B)+AB)z=0$, where $c=1-\lambda^{-1}$. 


Denote $Bz=u$, where $u$ may be arbitrary vector such that $(u,z)>0$. Next, $A(cz+u)=-z-cu$ and this is possible whenever $0<(cz+u,-z-cu)=-c(z,z)-(1+|c|^2)(u,z)-{\bar c}(u,u)$. This is possible if and only if the real part of $c$ is negative, i.e., $\Re \lambda^{-1}>1$, $|\lambda-1/2|<1/2$.


