<s>[INST] Prove that $f(x) = \sqrt{x^2 + x}$ is uniform continuity at the interval $[0, \infty)$. Prove that $f(x) = \sqrt{x^2 + x}$ is uniform continuity at the interval $[0, \infty)$.
There is a hint to the exercise:
Prove that $\forall x,y \geq 1: \frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}} \leq 2$
How does the hint help me? 
 [/INST] $f$ is clearly uniformly continuous on $[0,1]$. 
To show uniform continuity on $[1, +\infty\rangle$ notice that for $x \ge 1$ we have
$$f'(x) = \frac{2x+1}{2\sqrt{x(x+1)}} \le \frac{2\cdot 1+1}{2\sqrt{1(1+1)}}= \frac{3}{2\sqrt2}$$
so for $x,y \ge 0$ the mean value theorem gives that there exists $\theta$ between $x$ and $y$ such that
$$|f(x) - f(y)| = |f'(\theta)||x-y| \le \frac{3}{2\sqrt2}|x-y|.$$
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