Let $H$ be a complex Hilbert space with inner product $\langle\cdot,\cdot\rangle$. We say $A\geq 0$ if $\langle Ax,x\rangle\geq0$ for all $x\in H$. If $A\geq 0$ and $0\leq B\leq C$, then does $0\leq \sqrt{B}A\sqrt{B}\leq \sqrt{C}A\sqrt{C}$?
Attempt to start:
This question amounts to the statement, if $\langle Ax,x\rangle\geq0$ does $$0\leq\langle \sqrt{B}x,A\sqrt{B}x\rangle\leq\langle\sqrt{C}x,A\sqrt{C}x\rangle.$$ This is true if $A$ commutes with $B$ and $A$ commutes with $C$, since then $\sqrt{A}$ commutes with $\sqrt{B}$ and $\sqrt{A}$ commutes with $\sqrt{C}$, and therefore $\sqrt{B}A\sqrt{B}= \sqrt{B}\sqrt{A}\sqrt{A}\sqrt{B}=\sqrt{A}\sqrt{B}\sqrt{B}\sqrt{A}=\sqrt{A}B\sqrt{A}$, and $\sqrt{A}B\sqrt{A}\leq\sqrt{A}C\sqrt{A}$ if $A\geq 0$ and $B\leq C$.
But does this hold in the general setting?