Inequalities with single sided absolute values versus double sided absolute values I am trying to understand how the behavior between the absolute value(s) and the inequality sign differ between equations of the form:
\begin{equation}
|x| < a 
\tag{1}
\label{1}
\end{equation}
for some $a \in \mathbb{R}$ and
\begin{equation}
|f(x)| < |g(x)|
\tag{2}
\label{2}
\end{equation}
for $f : D_1 \subseteq\mathbb{R} \to \mathbb{R}$ and $g : D_2 \subseteq\mathbb{R}\to\mathbb{R}$.
Specifically, I know that when one solves an equation like eq. $\ref{1}$, you do the following:


*

*Solve $x < a$.

*Solve $x > -a$.

*Solve eq. $\ref{1}$ by taking the intersection of the solution found in steps 1. and 2.


I think of step 2. as first writing $-x < a$ (to make the LHS positive for the case when $x < 0$), and then multiplying both sides by $-1$, thereby flipping the inequality sign to get $x > -a$.
To solve eq. $\ref{2}$, you take each case separately, as shown in, for example, this question. However, in each of these cases, as far as I know, you do not switch the inequality sign like in step 2. for solving eq. $\ref{1}$. 
Therefore, my question is why, conceptually, is this done for solving eq. $\ref{1}$, but not done in each case (i.e. when the argument of one of the absolute values is negative) when solving eq. $\ref{2}$.
Thanks in advance. I have been avoiding absolute values + inequalities for most of my career in mathematics; it's time I face them.
 A: Note that the absolute value of a number equals its distance from $0$ which is always a non-negative number.
So the statement that $\vert x\vert<a$ tells us that $a$ is a positive number and that $x$ is closer to $0$ than $a$ is. From this we conclude that $x$ lies somewhere between $-a$ and $a$. In terms of inequalities
$$ -a<x\text{ and }x<a$$
A pair of inequalities connected by the word "and" [but not the word "or"] can be united in a combined inequality provided both inequalities are in the same direction. So the pair of inequalities above can be written in combined form as
$$-a<x<a$$
which is read as "negative $a$ is less than $x$ and $x$ is less than $a$" or more simply $x$ is between negative $a$ and positive $a$.
Now for the case where $|f(x)|<|g(x)|$.
Note that for non-negative numbers, the function $y=x^2$ is increasing. So it follows that  $0\le a<b$ if and only if $a^2<b^2$. Therefore
$$ |f(x)|<|g(x)|\text{ iff }f^2(x)<g^2(x) $$
which is equivalent to
$$ g^2(x)-f^2(x)>0 $$
which is equivalent to
$$ [g(x)-f(x)]\cdot[g(x)+f(x)]>0 $$
which is true if and only if $g(x)-f(x)$ and $g(x)+f(x)$ are both the same sign--both positive or both negative.
Ultimately, it means that $f(x)$ must always lie between $g(x)$ and $-g(x)$. That is
Either 


*

*$-g(x)<f(x)<g(x)$ or 

*$g(x)<f(x)<-g(x)$. 
So to find the solution set of $|f(x)|<|g(x)|$ you would find the solution of both 1 and 2 and take their union since the two inequalities are connected by the word "or."
