Subtracting 6 from both the sides, we get–

|x–1|+|x–2|+|x–3|–6≥0

Here, we have 4 cases:

Case 1: –∞ <x<1

For this case, |x–1|=–(x–1), |x–2|=–(x–2) and |x–3|=–(x–3)

⇒ –(x–1+x–2+x–3+6)≥0

⇒ x–1+x–2+x–3+6<0

⇒ 3x<0

⇒ x<0

⇒ xϵ (–∞ , 0) …(1)

Case 2:1<x<2

For this case, |x–1|=x–1, |x–2|=–(x–2) and |x–3|=–(x–3)

⇒ x–1–x+2–x+3–6≥0

⇒ –x–2≥0

⇒ x+2<0

⇒ x<–2

Which doesn’t signify the interval

Case 3:2<x<3

For this case, |x–1|=x–1, |x–2|=x–2 and x–3=–(x–3)

⇒ x–1+x–2–x+3–6≥0

⇒ x–6≥0

⇒ x≥ 6

Which doesn’t signify the interval

Case 4:3<x<∞

For this case, |x–1|=x–1, |x–2|=x–2 and |x–3|=x–3

⇒ x–1+x–2+x–3–6≥0

⇒ x–1+x–2+x–3–6≥0

⇒ 3x–12>0

⇒ x>4

⇒ xϵ (4 , ∞ ) …(2)

⇒ x ϵ (–∞ , 0)⋃ (4 , ∞ ) (from 1 and 2)

We can verify the answers using graph as well.