How do you prove this consequence of the triangle inequality?

[AxiomOfChoice]

I should know how to do this, but I just can't figure it out. Should be a piece of cake. How do you prove, for x,y\in \mathbb{R}^n:

<br /> \left| (||x|| - ||y||) \right| \leq ||x-y||<br />

 

[arildno]

We have:
||x+y||\leq{||}x||+||y||

Now, set x=z-y

Then, we get:
||z||\leq{||z-y||}+||y||,
For completely arbitrary z and y.

utilize this to derive the desired relation, i.e:
-||u-v||\leq{||u||}-||v||\leq{||}||u-v||

 

[poutsos.A]

I should know how to do this, but I just can't figure it out. Should be a piece of cake. How do you prove, for x,y\in \mathbb{R}^n:

<br /> \left| (||x|| - ||y||) \right| \leq ||x-y||<br />

||x||=||x-y+y||\leq||x-y||+||y|| ======>

||x||-||y||\leq||x-y||............1

||y||= ||y-x+x||\leq||x-y||+||x||======>

||y||-||x||\leq||x-y|| =======>

||x||-||y||\geq-||x-y||............2

from (1) and (2) we get:


-||x-y||\leq ||x||-||y||\leq ||x-y|| \Longleftrightarrow|(||x||-||y||)|\leq||x-y||