Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous.

So lets take:
${\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}^2 \, \leqslant \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}{\mid \sqrt{x^2+x} + \sqrt{y^2+y} \mid} \,\, = \,\, {\mid x^2+x-y^2-y \mid} \,\, = \,\, {\mid (x+y)(x-y)+(x-y) \mid} \,\, = \,\, {\mid (x-y)(x+y+1) \mid } < {\epsilon}^2 \Rightarrow \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid} < \epsilon$
So now we know that if we take $\delta = {\epsilon}^2$ the condition for uniform continuity of this function will be met because
${\mid x - y \mid} < (\delta = {\epsilon}^2) \Rightarrow {\mid f(x)-f(y) \mid} < \epsilon$
Is this proof valid? Or I miss something?

Hint: Multiply by conjugate:
$$\left|\sqrt{x^2+x}-\sqrt{y^2+y}\right|=\left|\sqrt{x^2+x}-\sqrt{y^2+y}\right|\frac{|\sqrt{x^2+x}+\sqrt{y^2+y}|}{|\sqrt{x^2+x}+\sqrt{y^2+y}|}$$
Note that this part of the proof only works with $[A,\infty)$, where $A > 0$. For the interval $[0,A]$, observe that $[0,A]$ is closed, and $f$ is continuous over that closed interval. There should be a theorem that says if $f$ is continuous over the closed interval then $f$ is uniformly continuous over that interval.
