Proving that for p.d. matrices A and B that $\|A(A+B)^{-1}\|_2<1$

Question

Let $A$ and $B$ be positive definite real matrices. It seems obvious to me that it should be true that

$$\|A(A+B)^{-1}\|_2<1.$$

My heuristic argument is that

$$\|A(A+B)^{-1}\|_2< \|A(A+\lambda_{\min}(B) I)^{-1}\|_2 = \max\left\{\frac{\lambda_i(A)}{\lambda_i(A)+\lambda_{\min}(B)}\right\}<1.$$

However, I'm not able to justify the first inequality.

Answer 1

This is actually false. If $\|x\|=1$, then $$ \|A(A+B)^{-1}x\|^2 = \langle A(A+B)^{-1} x, A(A+B)^{-1} x \rangle =\langle x, (A+B)^{-1}A^2(A+B)^{-1} x \rangle \\ = 1 - \langle y, ((A+B)^2 - A^2 ) y \rangle , $$ with $y=(A+B)^{-1}x$. So your claim is equivalent to $(A+B)^2-A^2$ being a positive definite matrix, and this will not follow from your assumptions.

The key fact is that the function $t\mapsto t^2$ is not operator monotone (there used to be a Wikipedia article on this, which I don't seem to be able to locate).