Solve : `|x| - |x-2| ge 1`

Answer

To solve the inequality \( |x| - |x-2| \geq 1 \), we will follow these steps:

### Step 1: Identify Critical Points
The critical points for the absolute value functions occur when the expressions inside the absolute values equal zero.

1. For \( |x| \):
- \( |x| = 0 \) when \( x = 0 \).

2. For \( |x-2| \):
- \( |x-2| = 0 \) when \( x = 2 \).

Thus, the critical points are \( x = 0 \) and \( x = 2 \).

### Step 2: Divide the Number Line into Intervals
We will divide the number line into three intervals based on the critical points:
1. \( (-\infty, 0) \)
2. \( [0, 2) \)
3. \( [2, \infty) \)

### Step 3: Analyze Each Interval

#### Interval 1: \( (-\infty, 0) \)
- Choose a test point, e.g., \( x = -1 \).
- Calculate \( |x| - |x-2| \):
\[ |x| = -x = 1 \quad \text{(since } x \text{ is negative)} \]
\[ |x-2| = |(-1)-2| = | -3 | = 3 \]
Thus,
\[ |x| - |x-2| = 1 - 3 = -2 \]
- Check the inequality:
\[ -2 \geq 1 \quad \text{(False)} \]
- Conclusion: No solutions in \( (-\infty, 0) \).

#### Interval 2: \( [0, 2) \)
- Choose a test point, e.g., \( x = 1 \).
- Calculate \( |x| - |x-2| \):
\[ |x| = 1 \quad \text{(since } x \text{ is positive)} \]
\[ |x-2| = |1-2| = 1 \]
Thus,
\[ |x| - |x-2| = 1 - 1 = 0 \]
- Check the inequality:
\[ 0 \geq 1 \quad \text{(False)} \]
- Conclusion: No solutions in \( [0, 2) \).

#### Interval 3: \( [2, \infty) \)
- Choose a test point, e.g., \( x = 3 \).
- Calculate \( |x| - |x-2| \):
\[ |x| = 3 \quad \text{(since } x \text{ is positive)} \]
\[ |x-2| = |3-2| = 1 \]
Thus,
\[ |x| - |x-2| = 3 - 1 = 2 \]
- Check the inequality:
\[ 2 \geq 1 \quad \text{(True)} \]
- Conclusion: Solutions exist in \( [2, \infty) \).

### Step 4: Combine Results
From our analysis:
- No solutions in \( (-\infty, 0) \).
- No solutions in \( [0, 2) \).
- Solutions in \( [2, \infty) \).

### Final Solution
The solution to the inequality \( |x| - |x-2| \geq 1 \) is:
\[ \boxed{[2, \infty)} \]