An inequality on matrix norm Does inequality $\|A\|_2\leq \|  |A|_m  \|_2 $  hold for all square matrices $A$ ? Where $|A|_m$ is also a square matrix, defined as $|A|_m := [|a_i,j|]$.
Two examples are provided for the case that the inequality holds.
(Example 1)  If $A=\pmatrix{1 & -1\cr -2 & 3\cr}, |A|_m =\pmatrix{1 & 1\cr 2 & 3\cr}$ and $\|A\|_2= \|  |A|_m  \|_2 =3.8643$ 
(Example 2) If $A=\pmatrix{-4 & 5 & -3 \cr -1 & -3 & 3 \cr -1 & -5 & -2}, |A|_m =\pmatrix{4 & 5 & 3 \cr 1 & 3 & 3 \cr 1 & 5 & 2}$ and $\|A\|_2 = 8.2575 \leq\|  |A|_m  \|_2 =9.6660 $ 

Let $y\in\mathbb{C}^n$ be such that
$$
\|A\|_2=\max_{x\neq 0}\frac{\|Ax\|_2}{\|x\|_2}=\frac{\|Ay\|_2}{\|y\|_2}.
$$
We show that
$$\tag{1}
\|A\|_2=\frac{\|Ay\|_2}{\|y\|_2}\leq\frac{\||A|z\|_2}{\|z\|_2} \quad \text{for $z=|y|$.}
$$
First note that for any $x\in\mathbb{C}^n$, we have
$$\tag{2}
\|x\|_2^2=\sum_{i=1}^n|x_i|^2=\sum_{i=1}^n|\;|x_i|\;|^2=\||x|\|_2^2
\quad\Rightarrow\quad\|x\|_2=\||x|\|_2.
$$
Next,
$$
\|Ax\|_2^2=\||Ax|\|_2^2=\sum_{i=1}^n\left|\sum_{j=1}^na_{ij}x_j\right|^2
\leq\sum_{i=1}^n\left|\sum_{j=1}^n|a_{ij}x_j|\right|^2
=\sum_{i=1}^n\left|\sum_{j=1}^n|a_{ij}||x_j|\right|^2=\||A||x|\|_2^2
$$
and hence
$$\tag{3}
\|Ax\|_2\leq\||A||x|\|_2.
$$
From (2) and (3), we obtain (1) by setting $x=y$ and then
$$
\|A\|_2=\frac{\|Ay\|_2}{\|y\|_2}\leq\frac{\||A|z\|_2}{\|z\|_2}
\leq\max_{x\neq 0}\frac{\||A|x\|_2}{\|x\|_2}=\||A|\|_2.
$$
P.S.: The inequality of course holds for rectangular matrices as well.
