Find values of $x$ that verify the inequality

Question

I have to find the values of $x$ that verify the inequality $\operatorname{arcsin}(\frac 1x)-\operatorname{arccos}(\frac 1x) \geq0$.

Let $f(x) = \operatorname{arcsin}(\frac 1x) - \operatorname{arccos}(\frac 1x).$

I differentiated $f(x)$ and I get that $f'(x)= \frac {-2}{\sqrt{x^{2}-1}}$, so $f(x)$ is strictly decreasing over $\mathbb{R}$, right? But:

$\operatorname{arcsin}(x):[-1,1]\to[-\frac \pi2,\frac\pi2]$ and $\operatorname{arccos}(x):[-1,1]\to[0,\pi]$.

Does this affect my inequality and, if so, how?

Answer 1

First consider $\arcsin(x)-\arccos(x)\ge0$.

For $\displaystyle x\in\left[\frac{1}{\sqrt{2}},1\right]$, $\displaystyle \arcsin(x)\ge\frac{\pi}{4}\ge \arccos(x)$

For $\displaystyle x\in\left[-1,\frac{1}{\sqrt{2}}\right)$, $\displaystyle \arcsin(x)<\frac{\pi}{4}< \arccos(x)$

So, $\displaystyle \arcsin\left(\frac{1}{x}\right)-\arccos\left(\frac{1}{x}\right)\ge0$ if and only if $\displaystyle \frac{1}{\sqrt{2}}\le \frac{1}{x}\le 1$, i.e., $$1\le x\le \sqrt{2}$$