prove that | |x| - |y| | greater than or equal to |x-y|

Question

a proof question : prove that the absolute value of x and of y are greater than or equal to the equation of x-y

Answer 1

(1) x,y≥0: |x|=x and |y|=y so ||x|-|y||=|x-y|

(2) x≥0, y<0, let y=-z²: |x|=x and |y|=z² so ||x|-|y||=|x-z²|; |x-y|=|x+z²|.

But x+z²>x-z², because z²>-z², therefore ||x|-|y||<|x-y|.

(3) x<0, let x=-z², y≥0: |x|=z² and |y|=y so ||x|-|y||=|z²-y|; |x-y|=|-z²-y|=z²+y.

But z²-y<z², so |z²-y|<z². z²+y>z², therefore  ||x|-|y||<|x-y|.

(4) x<0, let x=-w², y<0, let y=-z²: |x|=w² and |y|=z² so ||x|-|y||=|w²-z²|; |x-y|=|-w²+z²|=|w²-z²|.

But w²+z²>z²-w², w²>-w², therefore ||x|-|y||=|x-y|.

EXAMPLES

(a) x=1, y=2: ||x|-|y||=|1-2|=1; |x-y|=|1-2|=1, ||x|-|y||=|x-y|.

(b) x=1, y=-1: ||x|-|y||=|1-1|=0; |x-y|=|1+1|=2; 0<2 so ||x|-|y||<|x-y|.

(c) x=-1, y=1:  ||x|-|y||=|1-1|=0; |x-y|=|-1-1|=2; 0<2 so ||x|-|y||<|x-y|.

(d) x=-2, y=-1: ||x|-|y||=|2-1|=1; |x-y|=|-2+1|=1; ||x|-|y||=|x-y|.

PROOF THAT ||x|-|y||≤|x-y|, NOT ||x|-|y||≥|x-y|.