需要分段讨论
1、当x>=3时
|x-1|+|x-2|+|x-3|-2
=x-1+x-2+x-3-2
=3x-8>0
|x-1|+|x-2|+|x-3|-2>0
2、当2=|x-1|+|x-2|+|x-3|-2
=x-1+x-2+(-(x-3))-2
=x-2>=0
|x-1|+|x-2|+|x-3|-2>=0
3、当1<=x<2时
|x-1|+|x-2|+|x-3|-2
=x-1+(-(x-2))+(-(x-3))-2
=-x+2>0
|x-1|+|x-2|+|x-3|-2>0
4、当x<1时
|x-1|+|x-2|+|x-3|-2
=(-(x-1))+(-(x-2))+(-(x-3))-2
=-3x+4>0
|x-1|+|x-2|+|x-3|-2>0
所以对任意x都有
|x-1|+|x-2|+|x-3|-2>=0
|x-1|+|x-2|+|x-3|>=2