Is $|\langle Ax,x\rangle| \leq \langle |A|x,x\rangle$ in a complex Hilbert space?

Question

Let $H$ be a complex Hilbert space, $A$ a bounded operator and $|A|$ the unique positive square root of $A^*A$. In a comment in Bounded self adjoint operator can be written as difference of positive operators it is suggested that $|\langle Ax,x\rangle| \leq \langle |A|x,x\rangle$, and that this is a consequence of the Polarization identity. How?

I'm trying to show that $A$ has an absolutely convergent trace $\sum\langle Ae_i,e_i\rangle$ if $|A|$ has.

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

Answer 1

$|\langle Ax, x\rangle|\leq \langle |A|x,x\rangle$ is NOT true for a general bounded operator $A$.

Example. $A=\left(\begin{array} & 0 & 1\\ 0 & 0 \end{array}\right)$,$x=\left(\begin{array} & 2\\ 1 \end{array}\right)$

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We call $A$ is trace class if $|A|$ is, which is precisely the definition.