If $A$ and $B$ are real, symmetric, positive definite matrices, then prove that $A(A+B)^{-1}B$ is positive definite 
If $A$ and $B$ are real, symmetric, positive definite matrices, then prove that $A(A+B)^{-1}B$ is positive definite.

If $A$ and $B$ are positive definite, then $A+B$ is positive definite, and inverse of a positive definite matrix is positive definite. So $(A+B)^{-1}$ is positive definite. I don't know how to proceed after that.
Please anyone help me solve it. Thanks in advance. 
 A: It's true because $A(A+B)^{-1}B=\left[B^{-1}(A+B)A^{-1}\right]^{-1}=(B^{-1}+A^{-1})^{-1}$ and the sum or inverses of any symmetric positive definite matrices are symmetric positive definite.
A: Note that $M = A(A + B)^{-1}B$ is positive definite if and only if $M + M^T$ is symmetric and positive definite. Also, $(A + B)^{-1}$ is symmetric since it is the inverse of a symmetric matrix.  Now, 
Note that
$$
A(A + B)^{-1}B = \\
(A+B-B)(A+B)^{-1}B = \\
(A+B)(A+B)^{-1}B - B(A+B)^{-1}B = \\
B - B(A + B)^{-1}B = \\
B[B^{-1} - (A + B)^{-1}]B
$$
Now, it suffices to show that $B^{-1} - (A + B)^{-1}$ is positive semidefinite.  This proof is quick if you're familiar with the Loewner ordering, but one direct proof would be as follows.  Note that
$$
A \quad \text{is positive semidefinite} \iff\\
(A + B) - B \quad \text{is positive semidefinite} \iff\\
B^{-1/2}(A + B)B^{-1/2} - I \quad \text{is positive semidefinite} \iff\\
B^{-1/2}(A + B)B^{-1/2} \quad \text{is symmetric with eigenvalues greater than $1$} \iff\\
[B^{-1/2}(A + B)B^{-1/2}]^{-1} \quad \text{is symmetric with eigenvalues less than $1$} \iff\\
B^{1/2}(A + B)^{-1}B^{1/2} \quad \text{is symmetric with eigenvalues less than $1$} \iff\\
I - B^{1/2}(A + B)^{-1}B^{1/2}\quad \text{is positive semidefinite} \iff\\
B^{-1} - (A + B)^{-1} \quad \text{is positive semidefinite}
$$
A: Given $\rm A, B \succ \mathrm O_n$, we form the following block matrix
$$\begin{bmatrix} \mathrm A + \mathrm B & \mathrm A\\ \mathrm A & \mathrm A\end{bmatrix} = \begin{bmatrix} \mathrm I_n & \mathrm I_n\\ & \mathrm I_n\end{bmatrix} \underbrace{\begin{bmatrix} \mathrm B & \\ & \mathrm A\end{bmatrix}}_{\succ \mathrm O_{2n}} \begin{bmatrix} \mathrm I_n & \\ \mathrm I_n & \mathrm I_n\end{bmatrix} \succ \mathrm O_{2n}$$
and, using the Schur complement, we conclude that
$$\mathrm A - \mathrm A \left( \mathrm A + \mathrm B \right)^{-1} \mathrm A \succ \mathrm O_n$$
Note that
$$\mathrm I_n = \left( \mathrm A + \mathrm B \right)^{-1} \left( \mathrm A + \mathrm B \right) = \left( \mathrm A + \mathrm B \right)^{-1} \mathrm A + \left( \mathrm A + \mathrm B \right)^{-1} \mathrm B$$
and, hence,
$$\left( \mathrm A + \mathrm B \right)^{-1} \mathrm B = \mathrm I_n - \left( \mathrm A + \mathrm B \right)^{-1} \mathrm A$$
Left-multiplying both sides by $\rm A$,
$$\mathrm A \, \left( \mathrm A + \mathrm B \right)^{-1} \mathrm B = \mathrm A - \mathrm A \left( \mathrm A + \mathrm B \right)^{-1} \mathrm A \succ \mathrm O_n$$
