SOLUTION: Prove that: |x-y| is greater than or equal to ||x|-|y|| Thanks

I prefer working with ≦ than with ≧, so I will prove the equivalent
inequality:
                   
                   ||x|-|y|| ≦ |x-y|

Any number either equals to its absolute value or to the negative of its
absolute value, so  -|z| ≦ z ≦ |z| is true for any number z.

     Let z = x-y       also       let z = y

-|x-y| ≦ x-y ≦ |x-y|   and    -|y| ≦ y ≦ |y|

Add those two inequalities:

          -|x-y|-|y| ≦ x-y+y ≦ |x-y|+|y|

           -|x-y|-|y| ≦ x ≦ |x-y|+|y|
          
         -(|x-y|+|y|) ≦ x ≦ |x-y|+|y| 

                      |x| ≦ |x-y|+|y| 

That is true because -a≦b≦a is equivalent to |b|≦a             

Subtract |y| from both sides:

                 |x|-|y| ≦ |x-y| 
                      
--------------------------

Similarly we can interchange x and y in the above, and get

                |y|-|x| ≦ |y-x|

The right sides of both those inequalities are equal since

|y-x|=|-(x-y)|=|x-y|

      So we have |x|-|y| ≦ |x-y|

and also we have |y|-|x| ≦ |x-y|.

||x|-|y|| is either equal to |x|-|y| or to |y|-|x|,

and in either case it is ≦ |x-y|.

Therefore we have proved that

            ||x|-|y|| ≦ |x-y|

Edwin