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?
 A: For $\;x\in [1,\infty)\;$:
$$\left(\sqrt{x^2+x}\right)'=\frac{2x+1}{2\sqrt{x^2+x}}\le\frac{2x+1}{2x}=1+\frac1{2x}\stackrel{\text{Why?}}\le\frac32$$
Thus, having a bounded derivative makes $\;\sqrt{x^2+x}\;$ uniformly continuous in $\;[1,\infty]\;$, and being continuous in the bounded, closed interval $\;[0,1]\;$ it is unif. continuous there as well. Thus, it is unif. continuous in the whole non-negative interval.
A: Let's work with $f$ on an interval $[a,\infty[$, with $a>0$.
$$f'(x) = \frac{2x + 1}{2\sqrt{x^2 + x}} \le \frac{2x + 1}{2x} = 1 + \frac{1}{2x} \le 1 + \frac{1}{2a} $$
Now using MVT, $f$ becomes $(1 + \frac{1}{2a})$-Lipschitz on $[a,\infty[$. But Lipschitz $\implies$ Uniformly continuous.
For $[0,a]$, use the fact that this is a closed interval in $\mathbb R$, and that $f$ is continuous on this interval.
Thus $f$ is uniformly continuous on $[0,a]$ and $[a,\infty[$ each, hence on their union as well.
A: 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.
