Question 771814
<pre>
|x-1| + |x-2| + |x-4| &#8807; 6

There are 8 cases to consider

1. x-1 &#8807; 0, x-2 &#8807; 0, and x-4 &#8807; 0
2. x-1 &#8807; 0, x-2 &#8807; 0, and x-4 &#8806; 0
3. x-1 &#8807; 0, x-2 &#8806; 0, and x-4 &#8807; 0
4. x-1 &#8807; 0, x-2 &#8806; 0, and x-4 &#8806; 0
5. x-1 &#8806; 0, x-2 &#8807; 0, and x-4 &#8806; 0
6. x-1 &#8806; 0, x-2 &#8807; 0, and x-4 &#8806; 0
7. x-1 &#8806; 0, x-2 &#8806; 0, and x-4 &#8807; 0
8. x-1 &#8806; 0, x-2 &#8806; 0, and x-4 &#8806; 0
 
They simplify to

1. x &#8807; 1, x &#8807; 2, and x &#8807; 4 &#8608; x &#8807; 4
2. x &#8807; 1, x &#8807; 2, and x &#8806; 4 &#8608; 2 &#8806; x &#8806; 4
3. x &#8807; 1, x &#8806; 2, and x &#8807; 4 &#8608; contradiction
4. x &#8807; 1, x &#8806; 2, and x &#8806; 4 &#8608; 1 &#8806; x &#8806; 2
5. x &#8806; 1, x &#8807; 2, and x &#8806; 4 &#8608; contradiction
6. x &#8806; 1, x &#8807; 2, and x &#8806; 4 &#8608; contradiction
7. x &#8806; 1, x &#8806; 2, and x &#8807; 4 &#8608; contradiction
8. x &#8806; 3, x &#8806; 2, and x &#8806; 4 &#8608; x &#8806; 2
 
Four of those are obviously contradictions
So we have only four cases to consider: 

1. x &#8807; 1, x &#8807; 2, and x &#8807; 4 &#8608; x &#8807; 4
2. x &#8807; 1, x &#8807; 2, and x &#8806; 4 &#8608; 2 &#8806; x &#8806; 4
4. x &#8807; 1, x &#8806; 2, and x &#8806; 4 &#8608; 1 &#8806; x &#8806; 2
8. x &#8806; 3, x &#8806; 2, and x &#8806; 4 &#8608; x &#8806; 2

Case 1: x &#8807; 1, x &#8807; 2, and x &#8807; 4 &#8608; x &#8807; 4

1. x-1 &#8807; 0, x-2 &#8807; 0, and x-4 &#8807; 0
  
  |x-1| + |x-2| + |x-4| &#8807; 6 

becomes:

   x-1  +  x-2  +  x-4  &#8807; 6
                  3x-7  &#8807; 6
                    3x  &#8807; 13
                     x  &#8807; {{{13/3}}}

Case 2: x &#8807; 1, x &#8807; 2, and x &#8806; 4 &#8608; 2 &#8806; x &#8806; 4

2. x-1 &#8807; 0, x-2 &#8807; 0, and x-4 &#8806; 0

   |x-1| + |x-2| + |x-4| &#8807; 6 

becomes:

   x-1  +  x-2  -  x+4  &#8807; 6
                   x+1  &#8807; 6
                     x  &#8807; 5
                       

That contradicts x &#8806; 4 so case 2 is eliminated

Case 4: x-1 &#8807; 0, x-2 &#8806; 0, and x-4 &#8806; 0

 4. x &#8807; 1, x &#8806; 2, and x &#8806; 4 &#8608; 1 &#8806; x &#8806; 2

   |x-1| + |x-2| + |x-4| &#8807; 6 

beomes:

    x-1  -  x+2  -  x+4  &#8807; 6
                   -x+6  &#8807; 6
                     -x  &#8807; 0
                      x  &#8806; 0

That contradicts x &#8807; 1 so case 4 is eliminated

 
Case 8: x-1 &#8806; 0, x-2 &#8806; 0, and x-4 &#8806; 0

 4. x &#8806; 1, x &#8806; 2, and x &#8806; 4 &#8608; x &#8806; 1


   |x-1| + |x-2| + |x-4| &#8807; 6  

becomes:

        -x+1 - x+2 - x+4 &#8807; 6
                  -3x+7  &#8807; 6
                    -3x  &#8807; -1
                      x  &#8806; {{{1/3}}}

So the two cases that aren't self- contradictory are
cases 1 and 8. So the original inequality is true when

           x  &#8806; {{{1/3}}}  or when  x  &#8807; {{{13/3}}}

The solution is the union of the solutions for those two cases:

         (-&#8734;,{{{1/3}}}) U ({{{13/3}}},&#8734;)

Edwin</pre>