Let $|\cdot|$ be an absolute value on a field $K.$
Let $\alpha \in K^\times$ and $\varepsilon>0$ be given.
I am stuck trying to prove the following statement:
"There exists $\eta>0$ such that for all $\beta \in K^\times$ with $|\alpha-\beta|<\eta\,$ one has $|\alpha\beta^{-1}-1|<\varepsilon$"
So for any $\eta>0$ one has $$|\alpha-\beta|<\eta \; \Rightarrow \; |\alpha\beta^{-1}-1|<\eta|\beta|^{-1}$$ and so I want to choose my $\eta$ so that $$\eta<\varepsilon|\beta| \; \text{ for all } \; \beta \in K^\times \; \text{ such that } \; |\alpha-\beta|<\eta.$$ But now I'm stuck because I want a lower bound $|\beta|$ but the inequality is "the wrong way around".
Help greatly appreciated!
EDIT (following hint below): Choose $\eta>0$ such that $$\eta<\min(1,\varepsilon)\cdot\frac{|\alpha|}{2}.$$ Then, since $\eta<|\alpha|/2,$ one has $$|\beta|\geq |\alpha|-|\alpha-\beta|>|\alpha|-\frac{|\alpha|}{2}=\frac{|\alpha|}{2}$$ and so, since $\eta<\varepsilon|\alpha|/2,$ it follows that $$\varepsilon|\beta|>\eta.$$
Hint: Show there is an $\eta'$ such that for $|\alpha-\beta|<\eta'$, we have $|\beta|>|\alpha|/2$. Then choose $\eta < \min(\eta', \frac{1}{2}\epsilon |\alpha|)$.