Trace of Hermitian Positive Semidefinite Matrix

Question

Well, the question I want to ask is as follows.

Suppose A and B are Hermitian Positive Semidefinite (PSD) matrices, I wonder if it is possible to prove $Tr(A*(A+B)^{-1})\in (0,1]$ (if it is correct)?

If not, how about when A is of rank 1?

Thanks a lot!

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Above is the original question.

In another version, I want to prove $Tr(a_i a_i^H (AA^H))=1$ provided $A$ is non-singular. ($a_i$ is the $i$th column of $A$. The upper script H means conjugate transpose.)

Any idea?

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I modify the original question as: if A and B are hermitian PSD and (A+B) is nonsingular, how to prove the following?

$Tr(A*(A+B)^{-1})\le rank(A)$

Answer 1

I have come up with a proof for the new version (actually simpler version of the original problem), as follows.

$Tr(a_i a_i^H (AA^H)^{-1})=a_i^H (AA^H)^{-1}a_i$

Since A is nonsingular and we have $A^{-1}a_i=e_i$, where $e_i$ is a column of all zeros except the $i$th element. The reason is that $A^{-1} A=I$, thus $A^{-1}a_i$ is the $i$th column of $I$, the identity matrix.

Thus $a_i^H (AA^H)^{-1}a_i=e_i^H e_i=1$.