1

I'm attempting to prove that $|x|-|y| \le |x-y|$.

I've come up with the following proof.

The proof relies on these results obtained from previous exercises:

  1. $-|x| \le x \le |x|$
  2. ${|x-y|=|y-x|}$

Case 1. $x \le 0$, $y \le 0$. Here, we have $|x|-|y| = -x -(-y) = y-x$. Since $y-x \le |y-x| = |x-y|$, it follows that $|x|-|y| \le |x-y|$.

Case 2. $x \ge 0$, $y \le 0$. In this case, we have $|x|-|y| = x-(-y) = x+y$. Since, here, $y\le x$, it is the case that $x-y \ge 0$; thus, $|x-y| = x-y$. Observe that $y\le -y$. Adding $x$ to both sides of the inequality gives us $x+y \le x-y$. Hence, we have $|x|-|y| \le |x-y|$ as desired.

Case 3. $x\le0, y\ge 0$. Here, $|x|-|y| = -x-y$. Since, in this case, $x\le y$, we have $x-y \le 0$; thus, $|x-y| = -(x-y) = y-x$. Here, we have $-y \le y$. Adding $-x$ to both sides gives us $-x-y \le y-x$. Thus, $|x|-|y| \le |x-y|$.

Case 4. $x \ge 0, y \ge 0$. In this case, $|x|-|y| = x-y$. Since $x-y \le |x-y|$, it follows that $|x|-|y| \le |x-y|$ as desired.

I suspect there is a more straightforward or brief proof for this out there. Can someone point me in that direction?

Martin Sleziak
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Isaac Kleinman
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2 Answers2

5

If we know the triangle inequality: $|x+y|\leq |x|+|y|$ then $$|x|=|x-y+y|\leq |x-y|+|y|\iff|x|-|y|\leq|x-y|$$

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Use this: $$|x| - |y| = |x-y+y|-|y| \leq |x-y| + |y| - |y| = |x-y|$$ Using only one triangle inequality and working for any norm.

AlexR
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