Example 11. Solve the inequality

$$
\left|1-\frac{|x|}{1+|x|}\right| \geqslant \frac{1}{2}
$$
Solution. The domain of admissible values of this inequality consists of all real numbers. The inequality is equivalent to the combination of two systems:

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-134.jpg?height=178&width=880&top_left_y=2158&top_left_x=585)

Let's solve the first system:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ | \frac { 1 } { 1 + x } | \geqslant \frac { 1 } { 2 } , } \\
{ x \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \frac { 1 } { 1 + x } \geqslant \frac { 1 } { 2 } , } \\
{ x \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{2-1-x}{1+x} \geqslant 0, \\
x \geqslant 0
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 1 - x \geqslant 0 , } \\
{ x \geqslant 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \leqslant 1, \\
x \geqslant 0
\end{array} \Leftrightarrow 0 \leqslant x \leqslant 1\right.\right.
\end{aligned}
$$

Let's solve the second system. The first inequality of this system is equivalent to the inequality

$$
\left|\frac{1}{1-x}\right| \geqslant \frac{1}{2}
$$

If \( x < 0 \), then \( \left|\frac{1}{1-x}\right| = \frac{1}{1-x} \). If \( x > 0 \), then \( \left|\frac{1}{1-x}\right| = \frac{1}{x-1} \). Therefore, the second system is equivalent to the system

$$
\left\{\begin{array}{l}
\frac{1}{1-x} \geqslant \frac{1}{2} \\
x < 0
\end{array} \text{ or } \left\{\begin{array}{l}
\frac{1}{x-1} \geqslant \frac{1}{2} \\
x > 0
\end{array}\right.\right.
$$

The inequality \( \left|\frac{1}{1-x}\right| \geqslant \frac{1}{2} \) is equivalent to the inequality \( (x-1)^2 > x^2 \), i.e., \( x < \frac{1}{2} \).