Consider the change of basis $C^tA(I+BA)^{-1}C$ for $C=I+BA$. One gets $A+ABA$, which is positive semidefinite (sum of two positive semidefinite matrices).

The most difficult part is to show that $I+BA$ is invertible. If not, then let $v$ be in its kernel. By taking the scalar product with $Av$, one gets that $v$ must be in the kernel of $A$, which implies $v=0$.

Consider the change of basis $C^tA(I+BA)^{-1}C$ for $C=I+BA$. One gets $A+ABA$, which is positive semidefinite being a some of two positive semidefinite.

The most difficult part is to show that $I+BA$ is invertible. If not, then let $v$ be in its kernel. By taking the scalar product with $Av$, one gets that $v$ must be in the kernel of $A$, which implies $v=0$.

Consider the change of basis $C^tA(I+BA)^{-1}C$ for $C=I+BA$. One gets $A+ABA$, which is positive semidefinite being a some of two positive semidefinite.

The most difficult part is to show that $I+BA$ is invertible. If not, then let $v$ be in its kernel. By taking the scalar product with $Av$, one gets that $v$ must be in the kernel of $A$, which implies $v=0$.

Consider the change of basis $C^tA(I+BA)^{-1}C$ for $C=I+BA$. One gets $A+ABA$, which is positive semidefinite (sum of two positive semidefinite matrices).

The most difficult part is to show that $I+BA$ is invertible. If not, then let $v$ be in its kernel. By taking the scalar product with $Av$, one gets that $v$ must be in the kernel of $A$, which implies $v=0$.