# Question #11f91

Feb 12, 2015

$| 1 - \left(| x \frac{|}{1 + | x |}\right) | \ge \frac{1}{2} f \mathmr{and} x \epsilon \left[- 1 , + 1\right]$

First note that $\left(1 + | x |\right)$ will always be a positive value greater than (the positive value) $| x |$.
Therefore $\left(| x \frac{|}{1 + | x |}\right) |$ will always be less than $1$
and
$1 - \left(| x \frac{|}{1 + | x |}\right)$ will always be greater than $0$
(so we can ignore the "outside" absolute value bars).

$1 - \left(| x \frac{|}{1 + | x |}\right) \ge \frac{1}{2}$ can be simplified as

$\frac{1}{2} \ge | x \frac{|}{1 + | x |}$

$\rightarrow$ $1 + | x | \ge 2 | x |$

$\rightarrow$ $1 \ge | x |$

$\rightarrow$ $x$ must fall within the closed range $- 1$ to $+ 1$