## anonymous 5 years ago im trying to prove the inverse triangle inequality$||x|-|y||\leq|x-y|$like this:$x=x-y+y,$$|x|=|x-y+y|,$$|x|\leq|x-y|+|y|,$$|x|-|y|\leq|x-y|.$i get stuck here; dont know if its legit or not to take the absolute value of the LHS and preserve the inequality. help?

1. watchmath

your statement is equivalent to $$-|x-y|\leq |x|-|y|\leq |x-y|$$ now use triangle inequality

2. anonymous

im a little confused... how can i do that?

3. watchmath

Do you know that the inequality $$|x|\leq a$$ is equivalent to $$-a\leq x\leq a$$ ?

4. anonymous

i agree, but how can i apply the triangle inequality to the above system?

5. anonymous

i thought i could only apply it to an expression.

6. watchmath

ok, let's look at the first inequality $$-|x-y|\leq |x|-|y|$$. Can you rewrite the inequality so that there is no minus (but all plus) by moving around the expressions?

7. watchmath

Can you see that it is equivalent to $$|y|\leq |x|+|x-y|$$?

8. anonymous

$-|x-y|\leq|x|-|y|\implies|y|\leq|x|+|x-y|?$

9. watchmath

great!

10. watchmath

Now if we apply the triangle inequality to $$|y|=|(y-x)+x|$$ what do we get?

11. anonymous

$|y|\leq|y-x|+|x|$

12. watchmath

yes, but that is the same as $$|y|\leq |x|+|x-y|$$ right ? since |y-x|=|x-y|

13. anonymous

i agree.

14. watchmath

so if you want a more linear argument we can like this for the first inequality: $$|y|=|(y-x)+x|\leq |y-x|+|x|=|x|+|x-y|$$ It follows that $$-|x-y|\leq |x|-|y|$$

15. watchmath

you can do something similar to the 2nd one

16. anonymous

that makes sense. and with this, we can conclude that$||x|-|y||\leq|x-y|?$

17. watchmath

yes, if you prove two inequalities: $$-|x-y|\leq |x|-|y|$$ and $$|x|-|y|\leq |x-y|$$ you conclude the original inequality

18. anonymous

beautiful. i see it now. thank you very much for your time and help. i appreciate it a lot!

19. watchmath

you are welcome :)