Why does the second term require $|x|\leq|y|$ not $|x|\leq y$?

Question

The book says that the relation $\sqrt{x^2-y^2} + \arccos\dfrac{x}{y} = 0$ where $y \neq 0$ implies $\left|x\right| \geq \left|y\right|$ and $|x| \leq |y|$.

I understand that $x^2 - y^2 \geq 0$ from where $x^2 \geq y^2 \implies |x| \geq |y|$. But I cannot understand how to get the second constraint. What I can get is that $-1 \leq \dfrac{x}{y} \leq 1 \implies |x| \leq y$.

Answer 1

Given that you accept $-1 \le \frac xy \le 1$ from the $\arccos$ function you need $|x| \le |y|$ to satisfy it. Your version of $|x| \le y$ would fail for $x=-1,y=-2$ but $\arccos \frac {(-1)}{(-2)}$ is perfectly well defined.

Answer 2

$-1 \le \frac xy \le 1$ then if $y > 0$ then $-y \le x \le y \implies |x| \le y=|y|$.

And if $y < 0$ then $-y \ge x \ge y \implies |x| \le -y = |y|$.