Show that $\sqrt{x(x-1)}$ is uniformly continuous on $[1, \infty)$ Show that the function $f(x) = \sqrt{x(x-1)}$ is uniformly continuous on $[1, \infty)$.
Attempt:
\begin{align}
\left|\sqrt{x(x-1)} - \sqrt{y(y-1)}\right| &= \left|\frac{x^2 - x - y^2 - y}{\sqrt{x(x-1)} + \sqrt{y(y-1)}}\right| \\
&= \left(\frac{x + y + 1}{\sqrt{x(x-1)} + \sqrt{y(y-1)}}\right)|x-y| \\
&< \left(\frac{1}{\sqrt{x(x-1)} + \sqrt{y(y-1)}}\right)|x-y|
\end{align}
From here, I have no idea how to get rid of the denominator so that I will only be left with $|x-y|$. At this point, I don't even know if my approach is correct. Please give me some hint on how I should proceed. Thanks in advance!
 A: First notice that $\lim_{x\to1^{+}}f(x)=0$ clearly. So fix $\varepsilon>0$; $\exists\delta'>0$ such that $0<x-1<\delta', x\in[1,\infty)$ implies $|f(x)|<\varepsilon$.
Then consider the intervals $[1, 1+\delta'+1]$ and $[1+\delta',\infty)$.
On $[1+\delta',\infty)$, by notice that the coefficient in the bracket of your attempt is bounded above (though your coefficient seems to be incorrect, after correcting it still you can check it's bounded, for example by fixing an arbitrary $y$ to check it's always bounded in $x$ by 1), we know easily that $f(x)$ is Lipschitz on this interval, thus uniformly continuous on it; hence, $\exists\delta_{1}>0$ such that $|x-y|<\delta_{1}$ implies $|f(x)-f(y)|<\varepsilon$ for all $x,y\in[1+\delta',\infty)$.
On compact $[1, 1+\delta'+1]$, since $f(x)$ is continuous clearly, we know it's uniformly continuous on this interval. So $\exists\delta_{2}>0$ such that $|x-y|<\delta_{2}$ implies $|f(x)-f(y)|<\varepsilon$ for all $x,y\in[1, 1+\delta'+1]$.
Finally consider $\delta=min\{\delta_{1}, \delta_{2}, 1\}>0$; then for any $|x-y|<\delta$, we know $x$ and $y$ must be either both in $[1+\delta',\infty)$ or both in $[1, 1+\delta'+1]$; so $|f(x)-f(y)|<\varepsilon$, which shows the uniform continuity as desired.
A: Hint
You can use or prove the following general result

if $f: D \to \mathbb R$ is a continuous map defined on $D \subseteq
\mathbb R$  which is uniformly continuous on $D \setminus [a,b]$ where
$[a,b]$ is a closed bounded interval, then $f$ is uniformly continuous
on $D$.

The prove is fairly easy using Heine–Cantor theorem that you probably know.
You can apply that result to $f(x) = \sqrt{x(x-1)}$ with for example $a=1, b=2$ and follow on with the inequalities you mention in the question.
A: You can use the following
Lemma: Let $f, g: [1,\infty) \to \mathbb R$ two continuous functions such that $\lim_{x \to \infty} \lvert f(x) - g(x) \rvert = 0$. If $g$ is uniformly continuous, then also $f$ is uniformly continuous.
Proof: Let $\epsilon > 0$. Choose $\delta' > 0$ such that $\lvert x - y \vert < \delta'$ implies $\lvert g(x) - g(y) \rvert < \epsilon/3$. W.l.o.g. we may assume that $\delta' \le 1$. Choose $R \ge 1$ such that $\lvert f(x) - g(x) \rvert < \epsilon/3$ for $x  \ge R$. Since $f$ is uniformly continuous on the compact interval $[1,R+1]$, we find $\delta'' > 0$ such that $\lvert x - y \vert < \delta''$  implies $\lvert f(x) - f(y) \rvert < \epsilon$ whenever $x, y \in [1,R+1]$. Now let $\delta = \min (\delta',\delta'') \le 1$. Consider $x,y \in [1,\infty)$ such that $\lvert x - y \rvert < \delta$. If one of $x,y$ is contained in $[1,R]$, then both $x,y$ are contained in $[1,R+1]$ and therefore $\lvert f(x) - f(y) \rvert < \epsilon$. If both $x,y$ are contained in $[R,\infty)$, then
$$\lvert f(x) - f(y) \rvert = \lvert f(x) - g(x) + g(x) - g(y) + g(y) - f(y) \rvert \\ \le \lvert f(x) - g(x) \rvert + \lvert g(x) - g(y) \rvert +  \lvert f(y) - g(y) \rvert < \epsilon .$$
The uniform continuity of $f(x) = \sqrt{x(x-1)} = \sqrt{(x -\frac 1 2)^2 - \frac 1 4}$ follows from the Lemma by taking $g(x) = x - \frac 1 2$ which is clearly uniformly continuous. Note that with $z = x - \frac 1 2$ we have $\sqrt{z^2 - \frac 1 4} < z$ for all $z$ and
$$z - \epsilon \le \sqrt{z^2 - \frac 1 4}$$
for $z  \ge \frac{\epsilon^2 + \frac 1 4 }{2\epsilon}$.
