Solve absolute value inequality I have to show the inequality
$$
\left|\frac{1}{2 + a}\right| < 1.
$$
How do I do this?
I know that a fraction is less than 1 when the denominator is greater than the numerator, but I cannot just check if $2 + a > 1$ because of the absolute value sign.
Edit
If I use
$$
\left|\frac{1}{2 + a}\right| < 1 \Leftrightarrow -1 < \frac{1}{2 + a} < 1.
$$
I have to look at the inequalities separately, i.e. $\frac{1}{2+a} > -1 \Leftrightarrow 1 > -2 - a \Leftrightarrow a > -3$ and $\frac{1}{2+a} < 1 \Leftrightarrow 1 < 2+a \Leftrightarrow a > -1$.
Since $a > -1 > -3$, $a$ must just be greater than $-1$. But what about $a = -2$ which yields a zero in the denominator?
 A: Both sides are positive, so you can take their reciprocals (of course the 'less than' flips to 'greater than'):
$$\left|\frac 1{2+a}\right| < 1 \iff \frac 1{\left|\frac 1{2+a}\right|} = \frac {|2+a|}{|1|}= |2+a| > 1$$
That is equivalent to an alternative:
$$(2+a) < -1 \lor (2+a) > 1$$
which resolves to:
$$a < -3 \lor a > -1$$
Equivalently
$$a\in (-\infty, -3)\cup (-1,\infty)$$
EDIT in reply to the comment
No, $1/(2+a)>−1$ does not imply $a>−3$.
When you multiply both sides by $(2+a)$ you must consider the sign of the multiplicand term. If the term is negative, the direction of an inequality gets reversed. So you have two possible cases here:
$$\color{red}{1/(2+a) > -1} \quad |\,\times(2+a)$$
$$\begin{cases}1 > -1\times(2+a) & \text{ if}\ (2+a) > 0 \\
\qquad \text{or} \\
1 < -1\times(2+a) & \text{ if}\ (2+a) < 0 \end{cases}$$
This is equivalent to
$$1 > -2-a \ \text{and}\ 2+a > 0 \ \text{or} \ 1 < -2-a\ \text{and}\ 2+a < 0$$
$$a > -3 \ \text{and}\ a > -2 \ \text{or} \ a < -3\ \text{and}\ a <-2$$
Finally
$$\color{red}{a > -2 \ \text{or} \ a < -3}$$
Similary from the other inequality we get
$$\color{green}{1/(2+a)<1} \quad |\,\times(2+a)$$
$$\begin{cases}1 < 1\times(2+a) & \text{ if}\ (2+a) > 0 \\
\qquad \text{or} \\
1 > 1\times(2+a) & \text{ if}\ (2+a) < 0 \end{cases}$$
$$1 < 2+a \ \text{and}\ 2+a > 0 \ \text{or} \ 1 > 2+a\ \text{and}\ 2+a < 0$$
$$a > -1 \ \text{and}\ a > -2 \ \text{or} \ a < -1\ \text{and}\ a <-2$$
$$\color{green}{a > -1 \ \text{or} \ a < -2}$$
Together they make
$$(\color{red}{a > -2 \ \text{or} \ a < -3}) \ \text{and} \ (\color{green}{a > -1 \ \text{or} \ a < -2})$$
so
$$a < -3 \ \text{or} \ a > -1$$
