Thank you very much for the fruitful information that you provided to me. It was really interesting learning about Loewner's theorem and a great help for closing some remaining gaps in my thesis!

sorry maybe i formulated the constant to sloppy. We know $x^\top A x \le c^+ x^\top D x$ and $x^\top A^{-1} x \le \frac{1}{c^-} x^\top D^{-1} x$ holds. Doesn't the addition of both inequalities yield $x^\top (A+A^{-1}) x = x^\top A x + x^\top A^{-1} x \le c^+ x^\top D x + \frac{1}{c^-} x^\top D^{-1} x \le \max\{c^+, 1/c^-\} x^\top (D + D^{-1}) x$ ?

And let me come up with another question - i think that should be the last one! Is it possible to make a statement concerning the spectral equivalence estimates of the sum, e.g., something like $(A+A^{-1}) \le \max\{c^+, \frac{1}{c^+}\} (D + D^{-1})$ based on the bounds that we deduced so far??

ok, so I still have one question: in our case, $A$ and $B$ have eigenvalues in (0,\infty ). so the eigenvalues of $-A$ and $-B$ are in (-\infty ,0) and for $-A^{-1}$ and $-B^{-1}$ they are in in (-\infty ,0) and (-\infty ,0). How can I now deduce the statement using Proposition 2.2 since there it is neccessary to have the eigenvalues in the range $(-1,0)$ ?

Thank you for the hint. I have seen that remark, but in proposition 2.2 (to which they refer in that remark) they state that in that case, the eigenvalues of both A and D have to be in (-1,0). In our case, due to positive definiteness of A and D, this does obviously not hold. So the statement does not hold?

Thank you - again - for the very profound answer. I have never heard of this theorem before but obviously, it was exactly what I was looking for!