Inequality with absolute value

Solve the following inequality over the real domain: \[ \frac{|x+1|}{x-1}\ge x \]

Resolution
The expression is undefined for \(x=1\).
- \(x\le -1: \frac{-x-1}{x-1}\ge x \longrightarrow -x-1 \le x^2-x \longrightarrow x^2\ge -1 \longrightarrow x\in (-\infty,-1\rangle\)
- \(-1\le x< 1: \frac{x+1}{x-1}\ge x \longrightarrow x+1 \le x^2-x \longrightarrow x^2-2x-1\ge 0 \longrightarrow \)
  \( \longrightarrow x_{1{,}2}=\frac{2\pm\sqrt{4+4}}{2}=1\pm\sqrt{2} \longrightarrow x\in \left\langle -1, 1-\sqrt{2}\right\rangle\)
- \(x> 1: \frac{x+1}{x-1}\ge x \longrightarrow x+1 \ge x^2-x \longrightarrow x^2-2x-1\le 0 \longrightarrow x\in \left(1{,}1+\sqrt{2}\right\rangle\)
Result
The solution is \(x\in \left(-\infty,1-\sqrt{2}\right\rangle \cup \left(1{,}1+\sqrt{2}\right\rangle\).