# Use of the reverse triangle inequality in epsilon proof

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 $$\bar{x}$$.

• Please read tag descriptions before using them: per the description in the algebraic-geometry tag, that tag is inappropriate for this question. – KReiser Feb 27 at 20:44
• 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. – John Omielan Feb 27 at 21:07

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.

• Thank you, I don't know how I could not see this... Thanks! – Mathbeginner Feb 27 at 20:52

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

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

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

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

• makes perfect sense, thanks! – Mathbeginner Feb 27 at 20:53

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.