Question: Following a reference from “Elementos de Topología general” by Angel Tamariz and Fidel Casarrubias.

> 
> Let be $a\in\mathbb{R}\setminus\{0\}$ and we choose $\epsilon>0$. We define $\delta=\min\left\{\frac{|a|}2,\frac{\epsilon|a|^2}2\right\}$ and so we observe that if $y\in B(a,\delta)$, then 
>  $$
> \left|\frac{1}y-\frac{1}a\right|=\frac{|y-a|}{|y||a|}\le\frac{2}{|a|^2}|y-a|<\frac{2}{|a|^2}\frac{|a|^2\epsilon}2=\epsilon
> $$
> 
> 
> and so we conclude that $\frac{1}x$ is continuous.
> 
> 
> 

Unfortunately I don't understand why $\frac{|y-a|}{|y||a|}\le\frac{2}{|a|^2}|y-a|$, that is $\frac{1}{|y|}\le\frac{2}{|a|}$. In fact we have 
$$
|y-a|<\delta\le\frac{|a|}2\Rightarrow-2|a|<a-\frac{|a|}2<y<a+\frac{|a|}2<2|a|\Rightarrow|y|<2|a|\Rightarrow\frac{1}{|a|}<\frac{2}y
$$
that is in contradiction with $\frac{1}{|y|}\le\frac{2}{|a|}$.
Could someone help me, please?

Answer: 
Since $\lvert y-a\rvert<\frac{\lvert a\rvert}2$, we have$$\lvert y\rvert\geqslant\bigl\lvert\lvert a\rvert-\lvert y-a\rvert\bigr\rvert=\lvert a\rvert-\lvert y-a\rvert>\frac{\lvert a\rvert}2.$$So$$\frac{\lvert y-a\rvert}{\lvert y\rvert\lvert a\rvert}<\frac{\lvert y-a\rvert}{\lvert a\rvert}\times\frac2{\lvert a\rvert}=\frac2{\lvert a\rvert^2}\lvert y-a\rvert.$$
$$ |a| = |y|\leq |a-y| < \delta < \frac{|a|}{2}$$

Thus 

$$ |y| > \frac{|a|}{2}$$