# Inequalities involving absolute value

Let $x,y\in \Bbb R$.\begin{align} |x|-1\le|y| & \iff|x|-1\le y \text{ or }|x|-1\le -y \\ & \iff |x|-1\le y \text{ or } -|x|+1\ge y \\ & \implies |x|-1\le 1-|x| \\ & \implies |x|\le 1 \\ \end{align}

Is the above calculation correct? I have doubt about $|x|-1\le 1-|x|$ step, I mean, how does 'or' lead us there?

Also, how do we decide if 'and' is appropriate, or, 'or': $|x|\le2\iff x\le2 \text{ and } -x\le2$ while, $|x|\ge2\iff x\ge2\text{ or }-x\ge2$. Why does one have and while other have or?

• You have to change the inequality when multiply by -1. – Rodrigo Fontana May 21 '18 at 5:17
• it works like this: $|x-a|\leq b \iff -b \leq x-a \leq b$ and the inequality sign changes when you multiply the sides by a negative quantity. – Alvin Lepik May 21 '18 at 5:39
• @RodrigoFontana, so where have I not changed that? Also, how do we get $|x|-1\le 1-|x|$ from $|x|-1\le y \text{ or } -|x|+1\ge y$? – Silent May 21 '18 at 6:17

First off, I'd like to point out the correct approach for the second part. Think carefully about what the following means: $$|x|\le 2$$ You can read it aloud, too. It says that the number $x$, taken with no sign, is smaller than $2$. You can clearly see that this is equivalent to: $$- 2 \le x \le 2$$ This means $x$ must be greater than minus two and lesser than two. If you considered all numbers greater than minus two or lesser than two, any number would go (since any number is either greater than minus two or lesser than two, or in between).
Now look at the other condition. $$|x|\ge 2$$ It says that the number $x$, taken with no sign, is bigger than $2$. You can clearly see that this is equivalent to: $$x \le - 2 \quad \text{or} \quad x \ge 2$$ There is no way other than using or here. Why is the or necessary here? Well, think about any number that is simultaneously less than minus two and bigger than two...
About your inequality, think about it as you would with any function. First, express $y$ as a function of $|x|$, and then go straight to listing all possible cases as $x$ varies. Can you take it from here?
• So, following your ideas, from ($|x|-1\le y$ or $-|x|+1\ge y$), I get ($|x|\le y+1$ or $-|x|\ge y-1$), and from that, i get ($-y-1\le x\le y+1$ or $-1+y\le x\le 1-y$). So, can we get from that $|x|\le 1$? By the way, thank you very much for elaborate answer. – Silent May 21 '18 at 11:30
• So, how do we get $|x|\le 1$? – Silent May 21 '18 at 11:36
• Sadly, you cannot infer that the modulus of $x$ is always less than one. In fact, if I were to take $y=5$, $x$ could range between $-6$ and $6$. The first two implications you have written in the first comment are correct. – Niki Di Giano May 21 '18 at 11:37