Suppose $A$ is a compact operator on a complex Hilbert space $H$. Then $|A|$ is defined to be the unique positive $A\geq 0$ (i.e. $\langle x,|A|x\rangle\geq0$ for all $x\in H$) such that $|{A}|^2=A^*A$.
Is it true that $A\leq |A|$? That is, $$\langle x,Ax\rangle\leq \langle x,|A|x\rangle$$ for all $x\in H$.
This is true if $A$ is self-adjoint, since by the spectral theorem there exists an orthonormal basis $(\phi_n)$ of eigenvectors of $A$ with corresponding eigenvalues $\lambda_n$ such that $Ax=\sum_n\lambda_n\langle \phi_n,x\rangle\phi_n$ for all $x\in H$. So $A^2x=\sum_n\lambda_n^2\langle \phi_n,x\rangle\phi_n$, and $|{A}|x=\sum_n|\lambda_n|\langle \phi_n,x\rangle\phi_n$.
Therefore $$\langle x,Ax\rangle=\langle x,\sum_k\lambda_k\langle \phi_k,x\rangle\phi_k\rangle=\sum_k\lambda_k|\langle \phi_k,x\rangle|^2\leq \sum_k|\lambda_k||\langle \phi_k,x\rangle|^2= \langle x,|A|x\rangle.$$
But what about the more general case of $A$ compact and not necessarily self-adjoint?
Let $H = \mathbb{C}^2$ and $ A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $. Then
$$ A^* A = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \quad\implies\quad |A| = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}. $$
So it follows that, for $\mathbf{x} = (x, y)$,
$$ \langle \mathbf{x}, A\mathbf{x}\rangle = \bar{x} y \qquad\text{and}\qquad \langle \mathbf{x}, |A|\mathbf{x}\rangle = |y|^2. $$
Now it is clear that
$$\langle \mathbf{x}, |A|\mathbf{x}\rangle - |\langle \mathbf{x}, A\mathbf{x}\rangle| = |y|(|y| - |x|) $$
can assume both positive and negative values. (Also, note that if $A$ is not self-adjoint, then $\langle \mathbf{x}, A\mathbf{x}\rangle$ need note be real-valued at all.)