Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

i thought i could only apply it to an expression.

Answer 6

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?

Answer 7

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

Answer 8

\[-|x-y|\leq|x|-|y|\implies|y|\leq|x|+|x-y|?\]

Answer 9

great!

Answer 10

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

Answer 11

\[|y|\leq|y-x|+|x|\]

Answer 12

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

Answer 13

i agree.

Answer 14

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

Answer 15

you can do something similar to the 2nd one

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

you are welcome :)