Question

If x satisfies $|x-1|+|x-2|+|x-3|\ge 6$ then prove that all numbers x which satisfy the above relation are given by $x\le 0$ or $x\ge 4$

Answer 1

Here the change point are x = 1 , 2 , 3
Hence we consider the following case
(I) x < 1
(II) 1 < x < 2
(III) 2 < x < 3
(IV) x > 3
case (I) x < 1
-(x - 1) - ( x - 2) - (x - 3) $\ge$ 3
-3x + 6 $\ge$6 or -3x $\ge$ 0 $\therefore$ x $\ge$\, 0
Which is < 1 and hence the solution
case (II) 2 $\ge$ x <3
(x - 1) - ( x - 2) - (x - 3) $\ge$ 3
-3x + 6 $\ge$6 or -x $\ge$ 2 x $\ge$\, -2
This does not satisfy given condition of case (II) Hence no solution
case (III) 2 $\ge$ x <3
(x - 1) - ( x - 2) - (x - 3) $\ge$ 3
x $\ge$\, 6
This does not satisfy given condition of case (III) Hence no solution
case (IV) x $\ge$3
(x - 1) - ( x - 2) - (x - 3) $\ge$ 3
x $\ge$\, 14 or x $\ge$\, 4
This does not satisfy given condition of case (III) Hence no solution
Thus the required solution by case I are IV are x $\ge$\, 0 or x $\ge$\, 4