TITLE: Norm estimation of identity plus two non-commuting self-adjoint operators
QUESTION [6 upvotes]: This is a problem that I have been stuck for a few months.
Let $X$ be a Hilbert space and $A:B:X\to X$ be two non-commuting semi-positive self-adjoint bounded linear operators. Is it true that
$$\|(I+A+B)^{-1}A\|\le 1.$$
 If it is, can you suggest any reference to me?

I was able to show $\|(I+A+B)^{-1}A\|\le \|A\|$ and $\|(I+A)^{-1}A\|\le 1$. However, I don't know how to show the desired result without assuming $\|A\|\le 1$ or $AB=BA$. 
If it helps, I can show $A=C^*D_1^*D_1C$ and $B=C^*D_2^*D_2C$ for some linear bounded $C,D_1,D_2$.

REPLY [6 votes]: Though two answers are already posted, let me explain how to understand that it is false from general reasoning. The inequality $\|(I+A+B)^{-1}A\|\leqslant 1$ is equivalent to $\|(I+A+B)^{-1}A x\|\leqslant \|x\|$ for any vector $x$. Take $x=A^{-1}(I+A+B)y$, the inequality rewrites as $\|y\|\leqslant \|y+A^{-1}(I+B)y\|$. We are given only that $A,B$ are positive definite, thus for given vector $z$, $A^{-1}z$ may be any vector which forms an acute angle with $z$. For $z=(I+B)y$ this in general allows the vector to form an obtuse angle with $y$ (unless $y$ is an eigenvector of $B$.) So, if the direction of $A^{-1}(I+B)y$ forms an obtuse angle with $y$ and $A^{-1}$ is small enough, the inequality $\|y\|\leqslant \|y+A^{-1}(I+B)y\|$ fails.