anonymous
  • anonymous
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?
Mathematics
chestercat
  • chestercat
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watchmath
  • watchmath
your statement is equivalent to \(-|x-y|\leq |x|-|y|\leq |x-y|\) now use triangle inequality
anonymous
  • anonymous
im a little confused... how can i do that?
watchmath
  • watchmath
Do you know that the inequality \(|x|\leq a\) is equivalent to \(-a\leq x\leq a\) ?

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anonymous
  • anonymous
i agree, but how can i apply the triangle inequality to the above system?
anonymous
  • anonymous
i thought i could only apply it to an expression.
watchmath
  • 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?
watchmath
  • watchmath
Can you see that it is equivalent to \(|y|\leq |x|+|x-y|\)?
anonymous
  • anonymous
\[-|x-y|\leq|x|-|y|\implies|y|\leq|x|+|x-y|?\]
watchmath
  • watchmath
great!
watchmath
  • watchmath
Now if we apply the triangle inequality to \(|y|=|(y-x)+x|\) what do we get?
anonymous
  • anonymous
\[|y|\leq|y-x|+|x|\]
watchmath
  • watchmath
yes, but that is the same as \(|y|\leq |x|+|x-y|\) right ? since |y-x|=|x-y|
anonymous
  • anonymous
i agree.
watchmath
  • 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 \(-|x-y|\leq |x|-|y|\)
watchmath
  • watchmath
you can do something similar to the 2nd one
anonymous
  • anonymous
that makes sense. and with this, we can conclude that\[||x|-|y||\leq|x-y|?\]
watchmath
  • watchmath
yes, if you prove two inequalities: \(-|x-y|\leq |x|-|y|\) and \(|x|-|y|\leq |x-y|\) you conclude the original inequality
anonymous
  • anonymous
beautiful. i see it now. thank you very much for your time and help. i appreciate it a lot!
watchmath
  • watchmath
you are welcome :)

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