Proving |abs(x)-abs(y)|≤|x-y|

[lilli-baba]

Homework Statement

already proved abs(x)-abs(y)≤abs(x-y)
I just don't know how to prove it

Homework Equations

The Attempt at a Solution

we have :abs(x)-abs(y)≤abs(x-y)
abs(x)-abs(y)/abs(x-y)≤1 / abs(x-y) is diff than 0

proving abs(abs(x)-abs(y))/abs(x-y))≤1 :abs(x-y)diff than 0 then...
abs(abs(x)-abs(y)/abs(x-y))≤1 / abs(x-y)=abs(abs(x-y))

a<=abs(a)
a=abs(x)-abs(y)/abs(x-y)) :a≤1
abs(x)-abs(y)/abs(x-y))≤abs(abs(x)-abs(y)/abs(x-y))...
can't continue from here...I don't think this is the right way to do it
please any help.

 

[MathematicalPhysicist]

Well,

from |x| <= |x-y|+|y|
and |y| <= |y-x|+|x|

So you have: |x|-|y| , |y|-|x| <= |x-y|
Then because ||x|-|y|| equals one of them, what can you infer?

 

[HallsofIvy]

I would do this in four cases:
1) x> 0, y> 0
2) x> 0, y< 0
3) x< 0, y> 0
4) x< 0, y< 0

And each case would have two "subcases"
a) |x|> |y|
b) |x|< |y|

 

[Curious3141]

The way I did it was to take $\displaystyle ||x| - |y||.||x| + |y||$ and simplify it algebraically, ending up with $\displaystyle ||x| - |y|| = |x-y|.|\frac{x+y}{|x| + |y|}|$.

It should be quite trivial to observe that the factor on the right is always positive, and has a maximum value of 1, and that only occurs when x and y are of the same sign.