Solve $|x-1|+|x-2|+|x-3| \geq 6$
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
$x \in(-\infty, 0] \cup[4, \infty)$
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