Thread: Proving ||z|-|w|| <= |z-w|

hi, I stumbled upon a simple demonstration but i can´t solve it:

||z|-|w|| <= |z-w|

Can someone help me?

Pedro

Originally Posted by Pedro

hi, I stumbled upon a simple demonstration but i can´t solve it:

||z|-|w|| <= |z-w|

Can someone help me?

Pedro

This is the reverse triangle inequality.

ok, thank you!

This proof relies on this fact: $\displaystyle a \geqslant 0\;\& \, - a \leqslant b \leqslant a\, \Leftrightarrow \,\left| b \right| \leqslant \left| a \right|$.

So $\displaystyle \left| u \right| \leqslant \left| {u - w} \right| + \left| w \right|\, \Rightarrow \,\left| u \right| - \left| w \right| \leqslant \left| {u - w} \right|$

Likewise $\displaystyle \left| w \right| - \left| u \right| \leqslant \left| {w - u} \right| = \left| {u - w} \right|$

Thus we now have $\displaystyle - \left( {\left| {u - w} \right|} \right) \leqslant \left( {\left| u \right| - \left| w \right|} \right) \leqslant \left( {\left| {u - w} \right|} \right) $

Can you use the fact to finish?

i understood what is discussed above... but i would like know how to explain and admit that
-|a| ≤ a ≤ |a|

and

-|b| ≤ b ≤ |b|

can anybody help me to prove this?
all i know is that a negative is of course less than a positive number

Please start a new thread for any new question.

You said that you understand that $\displaystyle a \geqslant 0\;\& \, - a \leqslant b \leqslant a\, \Leftrightarrow \,\left| b \right| \leqslant \left| a \right| .$

If you do, can we say that $\displaystyle |a|\le ||a||~?$

If so then is it not true that $\displaystyle -|a|\le a \le |a|~?$

The fact is, that any nonnegative number is equal to its absolute value. So if $\displaystyle \displaystyle a \geq 0$, then $\displaystyle \displaystyle a = |a|$.

However, any negative number is less than its absolute value (since the absolute value is always nonnegative). So if $\displaystyle \displaystyle a < 0$, then $\displaystyle \displaystyle a < |a|$.

So that means for any $\displaystyle \displaystyle a$ that $\displaystyle \displaystyle a \leq |a|$.


A similar argument works the other way to show $\displaystyle \displaystyle -|a| \leq a$.

So that means $\displaystyle \displaystyle -|a| \leq a \leq |a|$.

thanks you! so same in the -|b| ≤ b ≤ |b|.