Does $|\langle x,Ax\rangle |=\langle x,|A|x\rangle$ for bounded operators on Hilbert space?

If $A$ is a bounded linear operator on a Hilbert space $H$ is it true that $|\langle x,Ax\rangle |=\langle x,|A|x\rangle$ for all $x\in H$? If not, can we at least establish inequality in one direction? Here $|A|$ is the unique positive square root of the positive operator $A^*A$. Hints or references are always appreciated.

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You cannot expect equality to hold in general. Let $\{ e_n \}_{n \in \mathbb Z}$ be an orthonormal basis of your Hilbert space. Consider the shift operator $Se_n = e_{n+1}$. Then $S^*S=1$ and $|S|=1$ (the identity operator. We then have $$0= |\langle e_1 , Se_1 \rangle | \neq \langle e_1, |S|e_1 \rangle =1.$$

This also rules out one possible direction of the inequality. To see that we cannot have an inequality in the other direction, let's look at some $2\times 2$-matrices.

Let $A= \left( \begin{array}{cc} 0 & 1 \\ 5 & 0 \end{array} \right)$ and then $|A|= \left( \begin{array}{cc} 5 & 0 \\ 0 & 1 \end{array} \right)$.

Let $x= \left( \begin{array}{c} 1 \\ 2 \end{array} \right)$.

Then $| \langle x , Ax \rangle | = 12$ but $\langle x, |A| x \rangle=9$.

Thus not only are the values not equal, there is no inequality between them (in either direction) that always holds.

What is the issue here? Take the polar decomposition of $A= U|A|$. Here $U$ is a partial isometry. Then we are comparing $| \langle x, A x \rangle | = |\langle U^* x, |A| x \rangle |$ with $\langle x, |A| x \rangle$.

Should there be an inequality between these two sides? Think of what happens if we fix $|A|$ and allow the partial isometry $U^*$ to be ANY partial isometry. We are then looking at the values obtained by taking the inner product of $|A|x$ with the elements of the (possibly quite large) set $\{U^*x \}$. Depending on $U^*$, this could be either larger or smaller than $\langle x, |A|x \rangle$.

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