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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.

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    $\begingroup$ 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
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    $\begingroup$ You divided by $-2$ but didn't change the inequality direction. $\endgroup$ – badatmath Dec 17 '18 at 20:33
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Since $$ xy\le|x||y|$$ then you have $$ -2|x||y|\le -2xy. $$

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  • $\begingroup$ Thankyou so much! $\endgroup$ – PolynomialC Dec 17 '18 at 20:33
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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 $$-|x-y|\leq|x|-|y|\leq|x-y|\implies \big||x|-|y|\big|\leq|x-y|$$

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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 $xy\leq |x||y|$.

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