Is it true that $|AQA|\leq A|Q|A$ if $A,Q$ are compact self-adjoint operators with $A\geq 0$?

Question

Is it true that $|AQA|\leq A|Q|A$ if $A,Q$ are compact self-adjoint operators with $A\geq 0$ on a complex Hilbert space $H$? That is, $$\langle x,|AQA|x\rangle\leq \langle x, A|Q|A x\rangle$$ for all $x\in H$, where $|Q|$ is the unique positive compact operator such that $|Q|^2=Q^*Q=Q^2$.

An attempt at starting. By the spectral theorem, $Q=UDU^*$ where $U$ is unitary and $D$ is diagonal consisting of the eigenvalues of $Q$. Then $|Q|=U|D|U^*$ where $|D|$ consists of the absolute values of the eigenvalues of $Q$.

Have no idea how to proceed!

Answer 1

Let $$A=\begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix},\quad Q=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$ Then $|Q|=I,$ hence $$A|Q|A=\begin{pmatrix} 4 & 0\\ 0 & 1 \end{pmatrix}$$ On the other hand $$AQA=\begin{pmatrix} 0 & 2\\ 2 & 0 \end{pmatrix}$$ thus $$|AQA|=2I=\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}$$ Therefore the inequality $|AQA|\le A|Q|A$ fails.