I just "proved" something ridiculous and can't find the fault in my logic. It's probably something really simple and obvious that I'm just overlooking, or maybe not because none of my friends can find the logic gap either.
Theorem: Let $f:U\to\Bbb{R}$ be a non-negative, non-decreasing, differentiable function where $U\subseteq\Bbb{R}$. If there exists an $\epsilon > 0$ such that for all $x,\ y\in U$ such that $x \ge y$, $f$ satisfies $$ f(x) - f(y) \le (x - y)^{1+\epsilon} $$ then $f(x)$ is a constant function.
Note: The $\epsilon$ in the theorem is constant. In the following "example", take $\epsilon = 1$ and everything should still hold.
Proof: $$ f'(x_0) = \lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x - x_0}\le\lim_{x\to x_0^+}\frac{(x-x_0)^{1+\epsilon}}{(x-x_0)}=\lim_{x\to x_0^+}(x-x_0)^{\epsilon}=0 $$ but since we assumed $f$ is non-decreasing, it's derivative is never negative. Thus we have $0 \le f'(x) \le 0$ which implies $f'(x) = 0$.
Now let $f(x) = \sqrt{x}$, with $U = \{x\ge 1\}$. Clearly${}^*$, $\sqrt{x} - \sqrt{y} \le (x - y)^{1+\epsilon}$, and $\sqrt{x}$ is non-negative, non-decreasing, and differentiable. The theorem above thus states that $f'(x) = 0$.
$*$ - First show $\sqrt{x} - \sqrt{y} \le x - y$, which is equivalent to showing that $\sqrt{x} - x \le \sqrt{y} - y$ which is true because $\sqrt{x} - x$ is a decreasing function on the interval $[1/4,\infty)$ and $x \ge y$.
$$\sqrt{x} - \sqrt{y} \le (x - y)^{1+\epsilon} \Leftrightarrow \frac{x-y}{\sqrt{x} + \sqrt{y}} \le (x - y)^{1+\epsilon}\Leftrightarrow 1 \le (x - y)^{\epsilon} (\sqrt{x} +\sqrt{y})$$
But $ 1 \le (x - y)^{\epsilon} (\sqrt{x} +\sqrt{y})$ cannot be true as the RHS goes to $0$ when $y \to x$.