Does an analog of the absolute value is greater than the real part inequality hold for square matrices? Does the following hold for any square matrix $A$, $(AA^*)^{1/2}\geq (A+A^*)/2$,
where the superscript $*$ denotes the Hermitian transpose. If not, does it hold for some types of matrices at least?
Proof/any comment would be appreciated.
 A: No. Taking a random example:
$$
A=\begin{bmatrix}1 & -1 \\ 0 & 0\end{bmatrix},
\quad
(AA^{*})^{1/2}=\begin{bmatrix}\sqrt{2} & 0 \\ 0 & 0\end{bmatrix},
\quad
\frac{1}{2}(A+A^*)=\frac{1}{2}\begin{bmatrix}2 & -1 \\ -1 & 0\end{bmatrix}.
$$
The difference
$$
(AA^{*})-\frac{1}{2}(A+A^*)
=
\begin{bmatrix}\sqrt{2}-1 & 1/2 \\ 1/2 & 0\end{bmatrix}
$$
is not semidefinite.
P.S.: Tons of other counter-examples can be generated by a simple Matlab code.
A=randn(2);
assert(all(eig((A*A')^(1/2)-0.5*(A+A'))>=0));

A: This inequality is true for normal $A$, i.e. $A$ that commute with $A^*$, because than $A$ and $A^*$ can be simultaneously diagonalized, and the inequality $|z|\geq\text{Re}(z)$ holds for the eigenvalues.
In general, non-commutativity gets in the way. Here is a general way to construct counterexamples. Set $|A|:=(AA^*)^{1/2}$, then the inequality is equivalent to $\text{Re}(Ax,x)\leq(|A|x,x)$ for all $x$. Pick $B=B^*\geq0$ and a normalized $x$, which is not an eigenvector of $B$, in particular $\|Bx\|\neq0$ . Let $y:=\frac{Bx}{\|Bx\|}$, then $y$ is also normalized and there is a unitary matrix $U$ such that $Ux=y$. Then $A:=BU$ is a counterexample.
The point is that $|A|=(AA^*)^{1/2}=(BUU^*B^*)^{1/2}=B$ because $U$ is unitary. Since $x$ is not an eigenvector of $B$ by Cauchy-Schwarz $(Bx,x)<\|Bx\|$, so $(Ax,x)=(BUx,x)=(Ux,Bx)=(y,Bx)=(\frac{Bx}{\|Bx\|},Bx)=\|Bx\|>(Bx,x)=(|A|x,x)$, contrary to the inequality (the left hand side is real by construction).
Together with polar decomposition this construction shows that any matrix, which is not a multiple of a unitary matrix, can be multiplied by a suitable unitary matrix on the right to produce a counterexample. Indeed, any $M=|M|V$ with unitary $V$, so take $B=|M|$ as above and set $A:=MV^{-1}U=BU$. All you need is that $|M|$ has a non-eigenvector, but if it does not then $|M|=cI$ and $M$ is a multiple of a unitary.
Moreover, if $A$ is non-singular the inequality $|(Ax,x)|\leq(|A|x,x)$ for all $x$ implies normality. I wonder if $\text{Re}(Ax,x)\leq(|A|x,x)$ is already enough. If so then a non-singular $A$ satisfies the inequality if and only if it is normal.
A: (Note: the question has been edited. This answer is aimed at a previous version in which $A$ was specified to be positive.)
Yes, in fact we have equality.
The key observation is that $A$ being positive implies that it is self-adjoint. Indeed, $X^*AX\in\mathbb{R}$ implies $X^*AX=(X^*AX)^* = X^*A^*X$, so that $(AX,X) = (X,AX)$ for all $X$, where $(\cdot,\cdot)$ is the standard Hermitian product (i.e. $(X,Y)=X^*Y$). Now the value of $(X,AY)$ for any $X,Y$ can be written down in terms of the values of $(X,AX)$ for various $X$ using the polarization identity (see remark 3.1 here). The same goes for $(AX,Y)$ in terms of $(AX,X)$, with the same identity, and it follows that $(AX,X)=(X,AX)$ for all $X$ implies $(AX,Y)=(X,AY)$ for all $X,Y$. Thus $A$ is self-adjoint, i.e. $A=A^*$.
$AA^*$ is a positive matrix, and by definition $(AA^*)^{1/2}$ is the unique positive matrix whose square is $AA^*$. But since $A$ is positive and (we have just shown) $A=A^*$, $A$ is a positive matrix whose square is $AA^*$, so it is the unique one. Thus $(AA^*)^{1/2}=A$.
