I have some suspicion that the following is true, though I am stuck in proving it;
A matrix $A$ is positve definite while $B$ is positive semidefinite, meaning that they are both square, symmetric and have nonnegative eigenvalues, in particular $A$ has all positive eigenvalues.
Assume that $M = A + B$, then $M$ is positive definite, so is $M^{-1}$. Then $AM^{-1}B + BM^{-1}A$ is symmetric, and I think that its eigenvalues are nonnegative. That would make $AM^{-1}B + BM^{-1}A$ positive semidefinite, which would make me happy!
Any ideas if it is possible to prove, or disproove?
Yes. For any $t>0$, define $B_t=B+tI$. Then $$ \begin{aligned} A(A+B)^{-1}B+B(A+B)^{-1}A &=\lim_{t\to0^+}\left[A(A+B_t)^{-1}B_t+B_t(A+B_t)^{-1}A\right]\\ &=\lim_{t\to0^+}2(A^{-1}+B_t^{-1})^{-1}\\ &\succeq0 \end{aligned} $$