Question

How to prove that abs(x-y) greater than or equal to abs (y)- abs(x)

Answer 1

You mean like this:
|x-y|>=|y|-|x|

Answer 2

Yes. I solved one similar with the x and y switched on the right side of the equation using the triangular inequality theorem but i can't get it to work for this one.

Answer 3

I will help you

Answer 4

Please do thanks !!!

Answer 5

|x|+|y-x|>=|x+y-x|=|y|
|y|+|x-y|>=|x+y-y|=|x|
Move |x| to the right side in the first inequality and |y| to the second inequality. We get:
|y-x|>=|y|-|x|
|x-y|>=|x|-|y|
From absolute value properties we know that |y-x|=|x-y| ;
and if z>=a and z>=-a=>z>=|a|
from this properties=>|x-y|>=|y|-|x|

Answer 6

This is the triangular inequality therom yes?
I'm a little confused about the part with z.
Are you saying z >=a and Z>=-a then -a>=z>=\[\left| a \right|\]

Answer 7

Sorry. It looks that you are saying -a is greater than z and z is greater than absolute value of a. And i'm a little confused.

Answer 8

Oh got it thanks!!!1

Answer 9

-a is lower than z where is the trouble

Answer 10

I read the inequalities wrong. So many >= got me confused. But I got it when I wrote it :D