Solve |x−1|+|x−2|+|x−3|≥6

Solution:

We have, $|x-1|+|x-2|+|x-3| \geq 6 \quad \ldots$ (i)

As, $|x-1|=\left\{\begin{array}{l}x-1, x \geq 1 \\ 1-x, x<1\end{array}\right.$

$|x-2|=\left\{\begin{array}{l}x-2, x \geq 2 \\ 2-x, x<2\end{array}\right.$ and

$|x-3|=\left\{\begin{array}{l}x-3, x \geq 3 \\ 3-x, x<3\end{array}\right.$

Now,

Case I : When $x<1$,

$1-x+2-x+3-x \geq 6$

$\Rightarrow 6-3 x \geq 6$

$\Rightarrow 3 x \leq 0$

$\Rightarrow x \leq 0$

So, $x \in(-\infty, 0]$

Case II : When $1 \leq x<2$,

$x-1+2-x+3-x \geq 6$

$\Rightarrow 4-x \geq 6$

$\Rightarrow x \leq 4-6$

$\Rightarrow x \leq-2$

So, $x \in \phi$

Case III : When $2 \leq x<3$,

$x-1+x-2+3-x>6$

$\Rightarrow x \geq 6$

So, $x \in \phi$

Case IV: When $x \geq 3$,

$x-1+x-2+x-3 \geq 6$

$\Rightarrow 3 x-6 \geq 6$

$\Rightarrow 3 x \geq 12$

$\Rightarrow x \geq \frac{12}{3}$

$\Rightarrow x \geq 4$

So, $x \in[4, \infty)$

So, from all the four cases, we get

$x \in(-\infty, 0] \cup[4, \infty)$