Determine the set of $x \in \mathbb{R}$ with $||x+1|-|x-1|| < 1$

Question

What is the right way to approach this in general, take the outer absolute Value and then go to the inner ones, consider the inner cases first..?

Should I consider cases for example that $|x+1|-|x-1| < 0$ and $|x+1|-|x-1| \ge 0$ or rather think about the x values that seem interesting, like $-1$ and $1$?

Answer 1

For $|x|>1$ the inequality has no solutions because we get $2<1$, which is impossible.

But for $|x|\leq1$, we obtain $|2x|<1$, which gives the answer: $$-\frac{1}{2}<x<\frac{1}{2}$$ id est, the set is $\left(-\frac{1}{2},\frac{1}{2}\right)$

Answer 2

Suppose $x \le -1$ then $x + 1 \le 0$ so $|x + 1| = -(x + 1)$. Similarly, $|x - 1| = -(x - 1)$. So if $x \le -1$ then $$ ||x + 1| - |x - 1|| = |-(x + 1) - (-(x - 1))| = |-x-1+x-1|=2. $$ Therefore if $x \le -1$ then $||x + 1| - |x - 1|| = 2 \not< 1$.

There are three cases:

• $x + 1 \le 0$ and $x - 1 \le 0$ (i.e. $x \le -1$)
• $x + 1 \ge 0$ and $x - 1 \le 0$ (i.e. $-1 \le x \le 1$)
• $x + 1 \ge 0$ and $x - 1 \ge 0$ (i.e. $1 \le x$)

I've done the first. You can do the other two.