## Inequalities Involution Absolutes

To solve inequalities involving a modulus, we have to be careful about removing the modulus sign, as there may be an unintended change of sign.
$\frac{3x-1}{\| x+1 \|}\gt 2$
.
The denominator is always positive, and the numerator is positive for
$3x-1 \gt 0 \rightarrow x \gt 1/3$
.
$x+1 \gt 0$
if
$x \gt -1$
which is the case if
$x \gt 1/3$
. We can write
$\frac{3x-1}{ x+1 } \gt 2, \: x \gt 1/3$

Numerator and denominator are both positive so
$3x-1 \gt 2(x+1)=2x+2 \rightarrow 3x-2x \gt 2+1 \rightarrow x \gt 3$
.
Suppose
$\| \frac{3x-1}{ x+1} \| \gt 2$
.
Removing the modulus sign forces us to take signs into account. Either numerator and denominator have the same sign:
$3x-1, \: x+1 \lt 0 \rightarrow x \lt 1/3, \: x \lt -1 \rightarrow x \lt -1$
or
$3x-1, \: x+1 \gt 0 \rightarrow x \gt 1/3, \: x \gt -1 \rightarrow x \gt1/3$

The numerator, denominator positive case is given above. Suppose numerator and denominator are both negative. Multiplying by a negative number reverses the inequality.
$3x-1 \lt 2(x+1)=2x+2 \rightarrow 3x-2x \lt 2+1 \rightarrow x \lt 3,$

But
$x \lt -1$
so
$x \gt 3$
or
$x \lt -1$
(1)
Suppose only one of denominator and numerator are positive. It must be the numerator, since if the numerator were positive (
$x \gt 1/3$
) and the denominator negative (
$x \lt -1$
) which is not possible. Hence
$-1 \lt x \lt 1/3$
(2) .Hence
$3x-1 \gt 2(x+1)=2x+2 \rightarrow 3x-2x \gt 2+1 \rightarrow x \gt 3,$
.
This is inconsistent with (2) so only (1) is possible,