The problem in the image asks to "find the set of real number solutions" for the inequality:

$\left| |x - 1| - |x + 1| \right| \leq 1$

Step-by-step solution:

1. Breaking the absolute values: We need to first handle the absolute values by breaking them into cases based on the critical points for $x$, which are $x = -1$ and $x = 1$ (because these are the points where the expressions inside the absolute values change signs).

2. Case 1: $x \geq 1$:

   • When $x \geq 1$, we have: $|x - 1| = x - 1 \quad \text{and} \quad |x + 1| = x + 1$ So the inequality becomes: $\left| (x - 1) - (x + 1) \right| \leq 1$ Simplifying inside the absolute value: $\left| x - 1 - x - 1 \right| = \left| -2 \right| = 2$ Now we check the inequality: $2 \leq 1$ This is false. Hence, there are no solutions in the region $x \geq 1$.

3. Case 2: $-1 \leq x < 1$:

   • In this interval, $x - 1 \leq 0$ and $x + 1 \geq 0$, so: $|x - 1| = 1 - x \quad \text{and} \quad |x + 1| = x + 1$ The inequality becomes: $\left| (1 - x) - (x + 1) \right| \leq 1$ Simplifying inside the absolute value: $\left| 1 - x - x - 1 \right| = \left| -2x \right| = 2|x|$ The inequality now becomes: $2|x| \leq 1$ Dividing both sides by 2: $|x| \leq \frac{1}{2}$ This implies: $-\frac{1}{2} \leq x \leq \frac{1}{2}$ Therefore, in this region, the solution is $x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$.

4. Case 3: $x < -1$:

   • When $x < -1$, both $x - 1 < 0$ and $x + 1 < 0$, so: $|x - 1| = 1 - x \quad \text{and} \quad |x + 1| = -(x + 1) = -x - 1$ The inequality becomes: $\left| (1 - x) - (-x - 1) \right| \leq 1$ Simplifying inside the absolute value: $\left| 1 - x + x + 1 \right| = \left| 2 \right| = 2$ Checking the inequality: $2 \leq 1$ This is false. Hence, there are no solutions in the region $x < -1$.

Final solution:

The solution to the inequality is the interval:

$x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$

Would you like more details, or do you have any further questions?

Here are five related questions you could explore:

1. What happens if we replace the inequality $\leq 1$ with $< 1$?
2. How would the solution change if the constants inside the absolute values were different?
3. What is the geometric interpretation of absolute value inequalities?
4. How would we solve $\left| |x - 2| - |x + 3| \right| \leq 2$?
5. Can this method be extended to solve inequalities with more absolute values?

Tip: Breaking absolute value expressions into cases based on critical points is a powerful technique for solving complex inequalities.