1. **Stating the problem:** Solve the inequality involving absolute values: $$|x-1| < |x+1|$$ and analyze the solution intervals. 2. **Recall the definition of absolute value:** For any real number $y$, $$|y| = \begin{cases} y, & y \geq 0 \\ -y, & y < 0 \end{cases}$$ 3. **Analyze the inequality:** $$|x-1| < |x+1|$$ means the distance from $x$ to 1 is less than the distance from $x$ to -1. 4. **Consider the critical points where expressions inside absolute values change sign:** These are at $x=1$ and $x=-1$. 5. **Split the real line into intervals based on these points:** - Interval 1: $x < -1$ - Interval 2: $-1 \leq x < 1$ - Interval 3: $x \geq 1$ 6. **Evaluate the inequality on each interval:** - For $x < -1$: $$|x-1| = -(x-1) = 1 - x$$ $$|x+1| = -(x+1) = -x -1$$ Inequality becomes: $$1 - x < -x -1$$ Simplify: $$1 - x < -x -1$$ Add $x$ to both sides: $$1 < -1$$ This is false, so no solutions here. - For $-1 \leq x < 1$: $$|x-1| = 1 - x$$ $$|x+1| = x + 1$$ Inequality: $$1 - x < x + 1$$ Simplify: $$1 - x < x + 1$$ Subtract 1 from both sides: $$-x < x$$ Add $x$ to both sides: $$0 < 2x$$ Divide both sides by 2: $$0 < x$$ So, $x > 0$ in this interval. Since interval is $-1 \leq x < 1$, solution here is $0 < x < 1$. - For $x \geq 1$: $$|x-1| = x - 1$$ $$|x+1| = x + 1$$ Inequality: $$x - 1 < x + 1$$ Simplify: $$x - 1 < x + 1$$ Subtract $x$ from both sides: $$-1 < 1$$ This is always true. So, solution here is $x \geq 1$. 7. **Combine all solutions:** $$x \in (0,1) \cup [1, +\infty) = (0, +\infty)$$ 8. **Check equality case:** $$|x-1| = |x+1|$$ Square both sides: $$ (x-1)^2 = (x+1)^2 $$ Expand: $$ x^2 - 2x + 1 = x^2 + 2x + 1 $$ Simplify: $$ -2x = 2x $$ $$ -4x = 0 $$ $$ x = 0 $$ 9. **Summary:** - Inequality $|x-1| < |x+1|$ holds for $x \in (0, +\infty)$. - Equality holds at $x=0$. **Final answer:** $$\boxed{x \in (0, +\infty)}$$