Inequalities with arctan I don't understand how to solve inequalities with arctan, such as:
$$\arctan\left(\frac{1}{x^2-1}\right)\ge \frac{\pi}{4} $$
If someone could solve this and give me a very brief explanation of what they did, I'd be thankful.

Here are the steps
$$\arctan\left(\frac{1}{x^2-1}\right)\ge \frac{\pi}{4} $$
$$\frac{1}{x^2-1}\ge \tan\left(\frac{\pi}{4}\right)$$
$$\frac{1}{x^2-1}\ge 1$$
$$\frac{1}{x^2-1}-1\ge 0$$
$$\frac{1-x^2+1}{x^2-1}\ge 0$$
$$\frac{2-x^2}{x^2-1}\ge 0$$
$$\frac{\left(\sqrt{2}-x\right)\left(x+\sqrt{2}\right)}{(x+1)(x-1)}\ge 0$$
Can you take it from here?
