`|x-1|+|x-2|+|x-3| ge 6`

Answer

To solve the inequality \( |x-1| + |x-2| + |x-3| \geq 6 \), we will analyze the expression based on the critical points where the absolute values change, which are \( x = 1 \), \( x = 2 \), and \( x = 3 \).

### Step 1: Identify the intervals
The critical points divide the number line into four intervals:
1. \( x < 1 \)
2. \( 1 \leq x < 2 \)
3. \( 2 \leq x < 3 \)
4. \( x \geq 3 \)

### Step 2: Solve for each interval

#### Case 1: \( x < 1 \)
In this interval, all expressions inside the absolute values are negative:
\[ |x-1| = -(x-1) = -x + 1, \quad |x-2| = -(x-2) = -x + 2, \quad |x-3| = -(x-3) = -x + 3 \]
Thus, the inequality becomes:
\[ (-x + 1) + (-x + 2) + (-x + 3) \geq 6 \]
Simplifying:
\[ -3x + 6 \geq 6 \]
Subtracting 6 from both sides:
\[ -3x \geq 0 \quad \Rightarrow \quad x \leq 0 \]
Since we are in the interval \( x < 1 \), this condition is satisfied for \( x \leq 0 \).

#### Case 2: \( 1 \leq x < 2 \)
In this interval, \( |x-1| \) is positive and the others are negative:
\[ |x-1| = x - 1, \quad |x-2| = -(x-2) = -x + 2, \quad |x-3| = -(x-3) = -x + 3 \]
The inequality becomes:
\[ (x - 1) + (-x + 2) + (-x + 3) \geq 6 \]
Simplifying:
\[ -x + 4 \geq 6 \]
Subtracting 4 from both sides:
\[ -x \geq 2 \quad \Rightarrow \quad x \leq -2 \]
This condition does not satisfy the interval \( 1 \leq x < 2 \). Thus, there are no solutions in this case.

#### Case 3: \( 2 \leq x < 3 \)
In this interval, \( |x-1| \) and \( |x-2| \) are positive, while \( |x-3| \) is negative:
\[ |x-1| = x - 1, \quad |x-2| = x - 2, \quad |x-3| = -(x-3) = -x + 3 \]
The inequality becomes:
\[ (x - 1) + (x - 2) + (-x + 3) \geq 6 \]
Simplifying:
\[ x - 1 \geq 6 \]
Adding 1 to both sides:
\[ x \geq 7 \]
This condition does not satisfy the interval \( 2 \leq x < 3 \). Thus, there are no solutions in this case.

#### Case 4: \( x \geq 3 \)
In this interval, all expressions are positive:
\[ |x-1| = x - 1, \quad |x-2| = x - 2, \quad |x-3| = x - 3 \]
The inequality becomes:
\[ (x - 1) + (x - 2) + (x - 3) \geq 6 \]
Simplifying:
\[ 3x - 6 \geq 6 \]
Adding 6 to both sides:
\[ 3x \geq 12 \quad \Rightarrow \quad x \geq 4 \]
This condition satisfies the interval \( x \geq 3 \).

### Step 3: Combine the solutions
From the analysis, we found:
- Solutions for \( x < 1 \) are \( x \leq 0 \).
- Solutions for \( x \geq 3 \) are \( x \geq 4 \).

Thus, the final solution set is:
\[ x \in (-\infty, 0] \cup [4, \infty) \]