How to solve this weird inequality? $\frac{x-1}{x+1} < x$
Thanks!
I did the following.
$\frac{x-1}{x+1} - x< 0 /-x$
$\frac{x-1 - x(x+1)}{x+1} < 0$
$\frac{-x^2-1}{x+1} < 0$
What to do next?

Note that the inequality can be written as
$$-(x^2 + 1) \cdot \frac{1}{x + 1} < 0$$
Now $x^2 + 1$ is always positive, so this is equivalent to studying
$$\frac{1}{x + 1} > 0$$
Can you solve this?
