Use of the reverse triangle inequality in epsilon proof

Question

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 $\bar{x}$.

Answer 1

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.

Answer 2

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

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

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

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

Answer 3

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.