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 $x,y\in \mathbb{R}^n$:

$$\left| (||x|| - ||y||) \right| \leq ||x-y||$$

 

[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 $x,y\in \mathbb{R}^n$:

$$\left| (||x|| - ||y||) \right| \leq ||x-y||$$

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