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I'm trying to show that $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$. A hint would be nice.

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What is that heart stands for? I am not able to understand. – srijan May 30 '13 at 6:43
Wikipedia: Reverse triangle inequality; ProofWiki: Reverse Triangle Inequality. – Martin Sleziak May 30 '13 at 6:45
@srijan: I think the elements in LHS is so eager to be less that the RHS ones. Wholeheartedly eager!. – Babak S. May 30 '13 at 6:46
Maybe this calls for the love triangle inequality. – Erick Wong May 30 '13 at 6:47
The same thing for absolute value was asked here: Reverse Triangle Inequality Proof. It is a special case of your question, but the proofs are very similar. – Martin Sleziak May 30 '13 at 6:56
up vote 4 down vote accepted

Observe that

$\lVert x \rVert = \lVert (x -y) +y \rVert \leq \lVert (x -y) \rVert + \lVert y \rVert$

which gives

$\lVert x \rVert - \lVert y \rVert \leq \lVert x -y \rVert$ ... $(1)$


$-(\lVert x \rVert - \lVert y \rVert ) \leq \lVert (y -x) \rVert = \lVert (x -y) \rVert $... $(2)$

From $(1)$ and $(2)$ result follows.

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Use triangle inequality and norm properties to show that $$\lVert x\rVert-\lVert y\rVert\le\lVert x-y\rVert$$ and $$\lVert y\rVert-\lVert x\rVert\le\lVert x-y\rVert$$

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How about applying the triangle inequality to $\parallel x - y + y \parallel$?

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Fastest way to get the proof done – LeGrandDODOM May 30 '13 at 6:55
@Gabriel: Half-done, anyway. – Cameron Buie May 30 '13 at 7:01
@CameronBuie in cases like these where I am not sure if something is homework, I'd rather give an idea than a solution. Well, a full solution anyway. – Alex Wertheim May 30 '13 at 7:03
$\frac{\text{self improvement}}{\text{homework}}$ – Trancot May 30 '13 at 7:44

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