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I have trouble with the last step of a proof of this inequality:

$||c-d|-|e-f||\leq|c-e|+|f-d|$ for arbitrary $c,d,e,f\in\mathbb{R}$

Step 1:

Use the triangle inequality $|a+b|\leq|a|+|b|:$

with $a=c-e,\ b=f-d,\ a+b=c-e+f-d=(c-d)-(e-f)$

such that $|(c-d)-(e-f)|\leq|c-e|+|f-d|$

Step 2:

Use $x\leq|x|:$

$|(c-d)-(e-f)|\leq||c-d|-|e-f||$

Now I'm stuck. How do I go from here?

Arthur
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1 Answers1

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This is an application of the reverse triangle inequality, which states that $$\lvert\lvert x\rvert-\lvert y\rvert\rvert \leq \lvert x-y\rvert$$ This follows almost directly from the triangle inequality (as you can see here). Combining this result with the triangle inequality, we have that $$\lvert\lvert c-d\rvert-\lvert e-f\rvert\rvert \leq \lvert (c-d)-(e-f) \rvert = \lvert (c-e)+(f-d) \rvert \leq \lvert c-e \rvert+\lvert f-d \rvert$$

Community
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Michael L.
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  • But that means I have a mistake in step 2. What is the mistake? – Arthur Oct 26 '16 at 21:02
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    Well, the inequality you've written down doesn't follow from $x \leq \lvert x\rvert$ at all. Starting from $\lvert \lvert c-d\rvert-\lvert e-f\rvert\rvert$, you could write down $$\lvert c-d\rvert-\lvert e-f\rvert \leq \lvert \lvert c-d\rvert-\lvert e-f\rvert\rvert$$ Or, starting from $\lvert (c-d)-(e-f)\rvert$, you could write down $$(c-d)-(e-f) \leq \lvert (c-d)-(e-f)\rvert$$ – Michael L. Oct 26 '16 at 21:04