1. ## Solving inequality

See this example:

$\displaystyle \frac{x - 1}{x + 1} \geq \frac{2x}{x - 1}$

We multiply both sides by $\displaystyle (x + 1) (x - 1)$

and get:

$\displaystyle (x - 1)^2 \geq 2x(x +1)$

Criteria 1:

x < -1
$\displaystyle (x - 1)^2 \geq 2x(x +1)$
$\displaystyle x^2 - 2x + 1 \geq 2x^2 + 2x$
$\displaystyle x^2 + 4x -1 \leq 0$

$\displaystyle x_{1,2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}$

$\displaystyle x \in [-2-\sqrt{5}, -2+\sqrt{5}]$ but since $\displaystyle x < -1$ we get $\displaystyle x \in [-2-\sqrt{5}, 1)$

Criteria 2:
$\displaystyle x \in (-1, 1)$

$\displaystyle (x - 1)^2\leq 2x(x +1)$ why less or equal??

from that we get: $\displaystyle x^2 +4x -1 \geq 0 \Rightarrow x \in [-2+\sqrt{5},1)$ (why not (-1, -2+\sqrt{5}] ?)

Criteria 3:
x > 1
$\displaystyle x^2 + 4x - 1 \leq 0 \Rightarrow no solution$

All together:

$\displaystyle x \in [-2-\sqrt{5},-1) U [-2+\sqrt{5},1)$

2. If $\displaystyle x \in (-1,1)$, then $\displaystyle (x - 1)(x + 1)$ is negative.

3. short and simple. thanks