Prove : | |x| - |y| | ≤ | x – y |?

Answer 1

| |x| - |y| | ≤ | x – y |
square both sides
| |x| - |y| |^2 ≤ | x – y |^2
x^2-2|x||y|+y^2 ≤ x^2-2xy+y^2
cancelling everything you can, you get that:
-2|x||y| ≤-2xy
dividing by -2
xy ≤|x||y| ,
which is true, since x≤|x| and y≤|y|

Answer 2

Use the triangle inequality, which states that |a+b| <= |a|+|b|.

In this case, it means that

|x| = |x+y-y| = |(x-y)+y| <= |x-y| + |y|, so |x| - |y| <= |x-y|.

Also,

|y| = |y+x-x| = |(y-x)+x| <= |x-y| + |x|, so |y| - |x| <= |x-y|.

Since both |x|-|y| and -(|x|-|y|) are <= |x-y|, it follows that
||x|-|y|| <= |x-y|, since the absolute value must be equal to one of these.