Write the solution set of the inequation

Solution:

We have:

$|x-1| \geq|x-3|$

$\Rightarrow|x-1|-|x-3| \geq 0$

The LHS of the inequation has two seperate modulus. Equating these to zero, we obtain $x=1,3$ as critical points.

These points divide the real line in three regions, i.e $(-\infty, 1],[1,3],[3, \infty)$.

CASE 1: When $-\infty

$\therefore|x-1|-|x-3| \geq 0$

$\Rightarrow-(x-1)-[-(x-3)] \geq 0$

$\Rightarrow-x+1+x-3 \geq 0$

$\Rightarrow-2 \geq 0$

But this is not possible.

CASE 2 : When $1 \leq x \leq 3$, then $|x-1|=x-1$ and $|x-3|=-(x-3)$

$\therefore|x-1|-|x-3| \geq 0$

$\Rightarrow x-1+x-3 \geq 0$

$\Rightarrow \Rightarrow 2 x \geq 4$

$\Rightarrow x \geq 2$

$\Rightarrow x \in[2, \infty)$

CASE 3 : When $3 \leq x<\infty$, then $|x-1|=x-1$ and $|x-3|=x-3$

$\therefore|x-1|-|x-3| \geq 0$

$\Rightarrow x-1-x+3 \geq 0$

$\Rightarrow 2 \geq 0$

This is true.

Hence, the solution to the given inequality comes from case $s 2$ and 3 .

$[2, \infty) \mathrm{U}[3, \infty)=[2, \infty)$

$\therefore x \in[2, \infty)$