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I'm new to analysis and trying to prove something about a converging series.

Now I want to get from $|x_{n}-\bar{x}| < \frac{|\bar{x}|}{2}$ to the following statement $|x_{n}| > \frac{|\bar{x}|}{2}$ using the reverse triangle inequality, but I just don't seem to get it right.

As for as my knowledge goes, the reverse inequality states that $||b|-|a|| \leq |b|-|a|$. Any suggestions on how to apply this?

PS: it is a bout a converging sequence $x_{n}$ with limit $\bar{x}$.

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  • $\begingroup$ Please read tag descriptions before using them: per the description in the algebraic-geometry tag, that tag is inappropriate for this question. $\endgroup$ – KReiser Feb 27 at 20:44
  • $\begingroup$ It's likely a typo, but perhaps you were having difficulty because you have the reverse inequality sign mixed around. Your text says $||b|-|a|| \leq |b|-|a|$ (which won't be true if $|b| \lt |a|$), while it should be $||b|-|a|| \geq |b|-|a|$ instead. Also, you don't need to use the absolute values of $a$ and $b$ on the LHS. $\endgroup$ – John Omielan Feb 27 at 21:07
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You're almost right there. Note that $$|\bar x| - |x_n|\le |x_n-\bar x| < \frac{|\bar x|}2$$ gives you exactly what you want.

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  • $\begingroup$ Thank you, I don't know how I could not see this... Thanks! $\endgroup$ – Mathbeginner Feb 27 at 20:52
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$$ |x_n| - |\overline{x}| \leq |x_n - \overline{x}| < \frac{|\overline{x}|}{2} $$ and:

$$ |\overline{x}| -|x_n| \leq |x_n - \overline{x}| < \frac{|\overline{x}|}{2} $$

by definition of $||x| - |y||$.Hence:

$$ |x_n| > |\overline{x}| - \frac{|\overline{x}|}{2} = \frac{|\overline{x}|}{2} $$

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  • $\begingroup$ makes perfect sense, thanks! $\endgroup$ – Mathbeginner Feb 27 at 20:53
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The 'reverse' triangle inequality states the following: for all $x,y\in\mathbb{R}$ \begin{align*} |x-y|\geq |x|-|y|. \end{align*} To see this, notice that \begin{align*} |x|-|y|=|x-y+y|-|y|\leq |x-y|+|y|-|y|=|x-y|, \end{align*} where the inquality follows from the normal triangle inequality. Now, if you have $|x_n-\bar{x}|<\frac{|\bar{x}|}{2}$ then this means that \begin{align*} |\bar{x}|-|x_n|\leq\frac{|\bar{x}|}{2}, \end{align*} and so by adding $|x_n|$ to both sides and subtracting $\frac{|\bar{x}|}{2}$ the result follows.

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