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I need to show that $||x|-|y||\le|x+y|\le|x|+|y|$.
I proved the right inequality, $|x+y|\le|x|+|y|$, and now I need to prove the left inequality, $||x|-|y||\le|x+y|$.

I though that I could do it by using the definition of the absolute value, which means that I need to prove that $-|x+y|\le|x|-|y|\le|x+y|$, but I'm not sure how to proceed from here. How can I prove the last inequality?

dfeuer
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Yossi
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2 Answers2

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$$\bullet\;\;|x|=|x+y+(-y)|\le|x+y|+|y|\implies \begin{cases}|x|-|y|\le|x+y|\\{}\\|y|-|x|\ge -|x+y|\;\end{cases}\;\implies\ldots$$

And remember that for $\,x,a\in\Bbb R\;,\;\;a\ge 0$ :

$$|x|\le a\iff -a\le x\le a$$

DonAntonio
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ok i think i got it.

i need to show that $||x|−|y||≤|x+y|$ which means that $-|x+y|\le|x|-|y|\le|x+y|$. which means that i need to show that $|y|-|x+y|\le|x|\le|x+y|+|y|$

in general $|x|=|x+y-y|=|x+y+(-y)|$ and $|y|=|x+y-x|=|x+y+(-x)|$

now we know that $|x+y|\le|x|+|y|$ so $|x+y+(-y)|\le|x+y|+|-y|=|x+y|+|y|$ and $|x+y+(-x)|\le|x+y|+|y|$

so we have $|x|\le|x+y|+|y|$ (which proves the right side) and $|y|\le|x+y|+|x|$ which means that $|y|-|x+y|\le|x|$ (which proves the left side) so we get $|y|-|x+y|\le|x|\le|x+y|+|y|$ which proves that $||x|−|y||≤|x+y|$

is everything ok here?

Yossi
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  • @ Yossi: A cursory reading yields no obvious errors, looks OK to me. The inequality is usually written in the (equivalent) form $\vert \vert x \vert - \vert y \vert \vert \le \vert x - y \vert$. But no matter. – Robert Lewis Sep 08 '13 at 21:20