Solve `|x-1|+|x-2| ge 4`

Explanation

Let `f(x)=|x-1|+|x-2|`

Hence, solutions is `x in (-oo,-1//2] cup [7//2,oo)`.
Alternative Method:
Consider ` |x-1|+|x-2| lt 4`
Now, `|(x-1)+(x-2)| le |x-1|+|x-2|`
` :. |(x-1)|+(x-2)| lt 4 `
` implies |2x-3| lt 4`
`implies -4 lt 2x-3 lt 4`
`implies -1 lt 2x lt 7`
`implies -1//2 lt x lt 7//2`
Therefore, for `|x-1|+|x-2| ge 4`, we have
`x in (-oo,-1//2] cup [7//2,oo)`.