Is there a function with this property?

Is there a real function over the real numbers with this property $\ \sqrt{|x-y|} \leq |f(x)-f(y)|$ ? My guess is no but can anyone tell me why? This came up as a question of one of my collegues and i cant give an answer.

Indeed, no such function $f$ is possible.

Let $Y$ be the image of $f$, and let $g$ be the inverse of $f$. Then as pointed out by Eric Nitardy, $ |g(x)-g(y)|\le(x-y)^2$ for all $x,y$ in $Y$.

From this inequality it now follows that $g(Y)$ has measure zero, which contradicts the fact that $g(Y)$ is the whole real line. To prove $g(Y)$ has measure zero, it suffices to show that for any integer $n$, $g(Y_n)$ has measure zero where $Y_n=Y\cap[n,n+1].$ But for any $N$, $Y_n$ is contained in a union of $N$ intervals of diameter $1/N$ whose images under $g$ have diameter not exceeding $1/N^2$.

There is such a function on the rationals:

Let $(q_0,q_1,...)$ be any enumeration of the rationals. For each $i \ge 0$, let $f(q_i) = q_r$, where $r$ is the smallest index that maintains the required property of $f$. That is, $r$ is the smallest index such that $|q_r - f(q_j)| \ge \sqrt{|q_i - q_j|}$ for all $j \lt i$.

Edited to add: What's more, given any infinite subset $X \subset \mathbb{N}$, there is such a function $f: \mathbb{Q} \rightarrow X$. Just let $f(q_i)$ be the smallest $n \in X$ such that $|n - f(q_j)| \ge \sqrt{|q_i - q_j|}$ for all $j \lt i$.

Observe that $f(x) = f(y)$ implies $x = y$, so if $f$ is continuous then it must be monotonic. Now we are done by a fairly simple pigeonhole argument: if, say, $|f(1) - f(0)| = d$, then at least one of $|f(1) - f((n-1)/n)|, |f((n-1)/n) - f((n-2)/n)|, ...$ is at most $\frac{d}{n}$, and taking $n$ large enough this leads to a contradiction.

I have no idea what happens if $f$ isn't required to be continuous. But the idea here is that the condition gets harder to satisfy the closer $x$ and $y$ are.

Whenever $f(x_0)$ and $f(x_1)$ fall in an interval $[a, b]$, we must have $$|x_0 - x_1| \le (f(x_0) - f(x_1))^2 \le (a - b)^{2}.$$ Therefore, the pre-image of any interval $[a,b]$ under $f$ is contained in some interval of length $(a-b)^2$. Now let $a_0=0$ and $a_k=\sum_{n=1}^{k}n^{-1}$ for $k=1,2,3...$ Since $a_k$ diverges, we have $\cup_k [a_k, a_{k+1}] = [0, \infty)$. So we can write $$ \begin{eqnarray} f^{-1}\left([0, \infty)\right) &=& f^{-1}\left(\cup_k [a_k, a_{k+1}]\right) \\ &=& \cup_k f^{-1}\left([a_k, a_{k+1}]\right) \\ &\subseteq& \cup_k I_k, \end{eqnarray} $$ where each $I_k$ is an interval of length $(a_{k+1} - a_{k})^2 = 1/(k+1)^2$. This countable union of intervals is measurable, and has measure $\le \sum_{k=0}^{\infty} 1/(k+1)^2 = \pi^2/6$. Similarly, we can show that $f^{-1}\left((-\infty, 0]\right)$ is also contained in a measurable set with measure $\le \pi^2/6$. Combining these results, we have $f^{-1}(\mathbb{R}) \subseteq Q$, where $Q$ is measurable with measure $\pi^2/3$. Clearly $Q \neq \mathbb{R}$ (since $\mathbb{R}$ has infinite measure); hence $f^{-1}(\mathbb{R}) \neq \mathbb{R}$, and $f$ cannot be defined over the entire real line.

The same argument with $a_k$ replaced by $\sum_{n=1}^{k}(n+\alpha)^{-1}$ can be used to make the measure of $Q$ arbitrarily small (by choosing $\alpha$ sufficiently large), and we conclude that the domain of $f$ cannot contain any set of positive measure.

I think you're right, there are indeed none. It's pretty easy to prove that $ \left |\frac{f(x+\varepsilon) - f(x)}{\varepsilon} \right | \geq \frac{1}{\sqrt{\left |\varepsilon \right |}} $, now think about what this means when $\varepsilon$ tends to zero.

I'm not 100% sure about the non-differentiable/non-continuous case though.