Is the absolute value on matrices well-defined with respect to PSD cone?

Question

Consider for two matrices $A, B$ we say $A \geq B $ if $ A - B$ is positive semidefinite. Also, there exists a definition of absolute value for a matrix $A$ that is $|A| = \sqrt{A^\dagger A}$.

I was studying Hermitian matrices(matrices like $A$ with complex entries such that $A^\dagger = A$) and faced with a problem, is the following statement true for two arbitrary Hermitian matrices? $$ A \geq B \geq -A \implies |A| \geq |B|$$

Note that if $A$ is a Hermitian matrix then $|A|$ is PSD and the eigenvalues of $|A|$ are equal to the absolute value of the eigenvalues of $A$.

Answer 1

The statement is false. Consider $$A=\begin{bmatrix} 6 & 0 \\ 0 & 2\end{bmatrix}, \quad B=\begin{bmatrix} 1 & 2 \\ 2 & 1\end{bmatrix}.$$

We have $$A+B = \begin{bmatrix} 7 & 2 \\ 2 & 3\end{bmatrix} \ge 0, \quad A-B = \begin{bmatrix} 5 & -2 \\ -2 & 1\end{bmatrix} \ge 0$$ so $A \ge B \ge -A$. On the other hand, we have $$|A| = A, \quad |B| = \sqrt{B^*B} = \sqrt{\begin{bmatrix} 5 & 4 \\ 4 & 5\end{bmatrix}} = \begin{bmatrix} 2 & 1 \\ 1 & 2\end{bmatrix}$$ and therefore $$|A|-|B| = A-|B| = \begin{bmatrix} 4 & -1 \\ -1 & 0\end{bmatrix} \not\ge 0$$ so $|A| \ge |B|$ is false.

Answer 2

No, because the weaker result $$ A\succeq\left\{\matrix{X\\ Y}\right. \succeq0 \ \text{ and }\ XY=YX=0 \ \Longrightarrow\ A\succeq X+Y $$ (with $B=X-Y$) doesn't hold. For a counterexample, consider $$ A=\pmatrix{2&\sqrt{2}\\ \sqrt{2}&2},\ X=\pmatrix{1\\ &0},\ Y=\pmatrix{0\\ &1} $$ (and $B=X-Y=\operatorname{diag}(1,-1)$).