Eigenvalues of $A(A+B)^{-1}$ Given a positive semidefinite matrix $A$ and a positive definite matrix $B$ of the same dimension. Can we show that each eigenvalue:
$$ \lambda\{A(A + B)^{-1}\} < 1$$
(in the scalar case, this is $\frac{a}{a + b} < 1$ when both $a$ and $b$ are positive which holds trivially) 
 A: We know that for square matrices $C$, $D$, both $CD$ and $DC$ have the same eigenvalues. 
So $A(A+B)^{-1}$ has the same eigenvalues as 
$$
(A+B)^{-1/2}A(A+B)^{-1/2},
$$
which is positive definite.
We have
\begin{align}
I-(A+B)^{-1/2}A(A+B)^{-1/2}&=\\
&=(A+B)^{-1/2}(A+B)(A+B)^{-1/2}-(A+B)^{-1/2}A(A+B)^{-1/2}\\
&=(A+B)^{-1/2}B(A+B)^{-1/2},\end{align}
which is positive definite. 
Thus all eigenvalues of $A(A+B)^{-1}$ are strictly less than one. 
Edit: here is a clarification of the method used, in case it is not obvious. If $I-X$ is positive definite and $X=VDV^*$ with $D$ diagonal, then 
$$
I-X=I-VDV^*=V(I-D)V^*.
$$
So the diagonal of $I-D$ consists of positive entries, thus showing that each eigenvalues of $X$ is in $(0,1)$.
A: A partial answer:
In the case of $B(A+B)^{-1} = (A+B)^{-1}B$, the following arguments holds:
$B$ is positive definite, therefore $A+B \succ 0$ and thus $(A+B)^{-1} \succ 0$. Also, $B(A+B)^{-1}\succ 0$ (Since both commute), since $B \succ 0$. Now $A(A+B)^{-1} +B(A+B)^{-1} = I$, thus $B(A+B)^{-1}\succ 0 \Rightarrow I-A(A+B)^{-1}\succ 0 \Rightarrow \lambda(I-A(A+B)^{-1}) > 0 \Rightarrow \lambda(A(A+B)^{-1})<1$. Maybe that helps for some ideas.
