Prove that $f(x)=\frac{1}{x}$ is uniformly continuous on $(\frac{1}{2},\infty)$
Attempt:
We must show that for any $\epsilon>0$ we can find a $\delta > 0$, such that if $|x-a|<\delta$ and $|\frac{1}{x}-\frac{1}{a}|<\epsilon$ for all $a \in (\frac{1}{2},\infty)$.
$\quad$Scratchwork for $\delta$:.
$\begin{align} \quad \quad |\frac{1}{x}-\frac{1}{a}|=|\frac{a-x}{xa}|&=\fbox{$\frac{1}{|xa|}$}|x-a|\\&=4|x-a|<\epsilon \\ &\Rightarrow|x-a|<\frac{\epsilon}{4} \end{align}$
$\quad \quad $ **In order to bound $\frac{1}{|xa|}$ (or $\frac{1}{xa}$; since $a,x>0$), I used the fact that $x,a>\frac{1}{2}$.
$\quad \quad $Which would result in $\frac{1}{|xa|}=\frac{1}{xa}=\frac{1}{\frac{1}{2}*\frac{1}{2}}=4$.
$\quad$Choose $\delta=\frac{\epsilon}{4}\\$
$\quad$Proof:. Let $\epsilon>0$ be given and let $\delta=\frac{\epsilon}{4}$. If $|x-a|<\delta$ we have:
$\begin{align} \quad \quad |\frac{1}{x}-\frac{1}{a}|&=\frac{1}{|xa|}|x-a|\\&\leq4|x-a|\\ &<4\delta=4\frac{\epsilon}{4}=\epsilon. \end{align}$
$\rule{18cm}{.03cm}$
Is it enough to use the fact that that $x,a>\frac{1}{2}$, in order to place bounds on the function in order to reach a $\delta$? Any guidance or advice to strengthen the integrity of this proof would be greatly appreciated.
Actually, here is a much faster way to prove this result. Indeed, $f$ is $\mathcal{C}^1$ and one has, over $\left(\frac{1}{2},+\infty \right)$, $$|f'(x)|=\left| \frac{1}{x^2}\right| \leq 4$$
so $f$ is $4-$lipschitz, so $f$ is uniformly continuous over $\left(\frac{1}{2},+\infty \right)$.