Solve : ` | x - 1 |+ |x -2| ge 3`

Explanation

To solve the inequality \( |x - 1| + |x - 2| \geq 3 \), we will consider different cases based on the values of \( x \) that affect the absolute values.

### Step 1: Identify critical points
The absolute values \( |x - 1| \) and \( |x - 2| \) become zero at \( x = 1 \) and \( x = 2 \). Therefore, we will consider the intervals created by these points:
1. \( x < 1 \)
2. \( 1 \leq x < 2 \)
3. \( x \geq 2 \)

### Step 2: Case 1: \( x < 1 \)
In this case, both \( x - 1 \) and \( x - 2 \) are negative, so we can rewrite the inequality as:
\[ -(x - 1) - (x - 2) \geq 3 \]
This simplifies to:
\[ - x + 1 - x + 2 \geq 3 \]
\[ 3 - 2x \geq 3 \]
Subtracting 3 from both sides gives:
\[ -2x \geq 0 \]
Dividing by -2 (and flipping the inequality) results in:
\[ x \leq 0 \]
Thus, for this case, the solution is \( x \leq 0 \).

### Step 3: Case 2: \( 1 \leq x < 2 \)
In this interval, \( |x - 1| \) is positive and \( |x - 2| \) is negative. We rewrite the inequality as:
\[ (x - 1) - (x - 2) \geq 3 \]
This simplifies to:
\[ x - 1 - x + 2 \geq 3 \]
\[ 1 \geq 3 \]
This statement is false, which means there are no solutions in this interval.

### Step 4: Case 3: \( x \geq 2 \)
In this case, both \( x - 1 \) and \( x - 2 \) are positive, so we rewrite the inequality as:
\[ (x - 1) + (x - 2) \geq 3 \]
This simplifies to:
\[ x - 1 + x - 2 \geq 3 \]
\[ 2x - 3 \geq 3 \]
Adding 3 to both sides gives:
\[ 2x \geq 6 \]
Dividing by 2 results in:
\[ x \geq 3 \]
Thus, for this case, the solution is \( x \geq 3 \).

### Step 5: Combine the solutions
From our cases, we have:
1. From Case 1: \( x \leq 0 \)
2. From Case 2: No solutions
3. From Case 3: \( x \geq 3 \)

Combining these results, we find the solution set:
\[ x \in (-\infty, 0] \cup [3, \infty) \]

### Final Answer
The solution to the inequality \( |x - 1| + |x - 2| \geq 3 \) is:
\[ x \in (-\infty, 0] \cup [3, \infty) \]