Solving an inequality involving the absolute value

Question

I want to find the $x \in \mathbb{R}$ with $\left|\frac{x^2}{x^2-1}\right| > \frac{1}{2}$. My current approach looks like this: $$\left|\frac{x^2}{x^2-1}\right|> \frac{1}{2} \\ \Leftrightarrow x^2 > \frac{1}{2}|x^2-1| \\ \Leftrightarrow x^2 > \frac{1}{2}|x + 1||x-1| $$

At this point im stuck. What is a good approach to take from here?

Answer 1

An inequality with absolute value is really two inequalities because of the piecewise definition of absolute value.

$x^2 > \frac{1}{2}|x^2-1|$

is the same as

$x^2 > \frac{1}{2}(x^2-1) >-x^2$

which you can probably solve.