Wondering where my logic is going wrong in this assignment:

Show that $||x|-|y|| \leq |x-y|$

Using the fact $||x|-|y||, |x-y| \geq 0$

It follows $(|x|-|y|)^2 \leq (x-y)^2$

Using the fact $|x|^2 = x^2$

$x^2 -2|x||y| +y^2 \leq x^2 -2xy +y^2$

Cancelling down:

$|xy| \leq xy$

Which I know is not true. Thanks for any input.

  • 2
    $\begingroup$ You are dividing at the last step by $-2$. This changes the $\le$ into a $\ge$, giving you a correct statement. $\endgroup$ – Crostul Dec 17 '18 at 20:32
  • 1
    $\begingroup$ You divided by $-2$ but didn't change the inequality direction. $\endgroup$ – badatmath Dec 17 '18 at 20:33

Since $$ xy\le|x||y|$$ then you have $$ -2|x||y|\le -2xy. $$

  • $\begingroup$ Thankyou so much! $\endgroup$ – PolynomialC Dec 17 '18 at 20:33

You divided by $-2$ but did not change the sign. Another approach.

Observe $$|x|=|x-y+y|\leq|x-y|+|y|\implies |x|-|y|\leq|x-y|$$

Simillarly we have $|y|-|x|\leq|x-y|$.

Hence $$-|x-y|\leq|x|-|y|\leq|x-y|\implies \big||x|-|y|\big|\leq|x-y|$$


First, in line 3 you are already using the fact that $||x|-|y||\leq|x-y|$.

However, assuming this from the start (as I am guessing you want to move into something you already proved and then work your way back) and following each of the steps you took, notice that between lines 6 and 8 you would have and intermediate step as follows: $-2|x||y|\leq-2xy$ or equivalently $-|x||y|\leq -xy$.

There, you multiply both sides of the inequality by $-1$, inverting the inequality $\bigl((a\leq b)\rightarrow (-b\leq -a)\bigr)$ ending up with $xy\leq |x||y|$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.