<s>[INST] 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?
 [/INST] 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.
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