How to prove that $A B A^* \leq \|B\| A A^*$ for operators A,B? Let $A$, $B$ bounded operators on a Hilbert space $H$. Further let $B$ be self-adjoint. 
Then we have that $A B A^* \leq \|B\| A A^*$.
I wanted to ask how to prove this inequality or where I can find a proof if it is a very common ineuqality.
The usual way to prove this is in two steps. First you prove that for any selfadjoint $B$, you have $\|B\|\,I-B\geq0$. This is easy, because $\|B\|\,I-B$ is selfadjoint and $$\sigma(\|B\|\,I-B)=\{\|B\|-\lambda:\ \lambda\in\sigma(B)\}.$$ As $|\lambda|\leq\|B\|$ for all $\lambda\in\sigma(B)$ and $\sigma(B)$ is real, we have $\sigma(\|B\|\,I-B)\subset[0,\infty)$. 

The second part is that if $S\geq0$, then $ASA^*\geq0$. This follows from writing $S=TT^*$ for some $T$, and then $$ASA^*=ATT^*S^*=(AT)(AT)^*\geq0.$$

Then $$A(\|B\|\,I-B)A^*\geq0,$$ so $\|B\|\,AA^*\geq ABA^*$.
Proof: For all $h \in H$, $\langle A^* BA h, h \rangle = \langle BA h, A h \rangle \leq \|B\| \cdot \langle A h, A h \rangle =   \langle \|B\|A^*A h, h \rangle$.