# Thread: fiddling with inequalities (quick question)

1. ## fiddling with inequalities (quick question)

$||x-z|+|y-z|| \ge |x-y|$

and then I switch the sign on the LHS, which is true:

$||x-z|-|y-z|| \le |x-y|$

or

$||x-z|-|y-z|| = |x-y|$

2. I suppose obviously the second one...

So how would I prove:

$||x-z|-|y-z|| \le |x-y|$ for all $x, y, z \in X$ ?

x-z=t

y-z=s

x-y=t-s

4. Originally Posted by MichaelMath
$||x-z|+|y-z|| \ge |x-y|$
Start with $|x|\le |x-y|+|y|$ implies $|x|-|y|\le |x-y|$

Likewise $|y|-|x|\le |y-x|=|x-y|$

So $-|x-y|\le |x|-|y|\le |x-y|$.

That implies $\left||x|-|y|\right|\le |x-y|$

5. Thanks mate, I figured it out using Zarathustra's suggestion. Can you get me started on this one?

$|d(x,z)-d(y,u)| \leq d(x,y)+d(z,u)$

6. Start by showing $d(x,z)-d(z,y)\le d(x,y)~\&~d(y,z)-d(y,u)\le d(z,u)$.

Note that $d(y,z)=d(z,y)$ and add those to together.

7. I need to read a bit about manipulating inequalities and absolute values... any suggestions?

8. Originally Posted by MichaelMath
I need to read a bit about manipulating inequalities and absolute values... any suggestions?
Can you get
$d(x,z)-d(y,u)\le d(x,y)+d(u,z)~\&~d(y,u)-d(x,z)\le d(u,z)+d(x,y)?$

Do you understand that if $a>0~\&~-a then $|b|< a?$

9. Okay...

$d(x,z)-d(z,y)\le d(x,y)$

$d(y,z)-d(y,u)\le d(z,u)$

So:

$
d(x,z)-d(z,y)+d(y,z)-d(y,u) \leq d(x,y)+d(z,u)$

$d(x,z)-d(y,u) \leq d(x,y)+d(z,u)$

and the second one...

10. ok, i got it.