If $|f(x) - f(y)| \leq C|x - y|^2$ for all $x, y \in (a, b)$, then dividing through by $|x-y|$ (for $x \neq y$) shows that $\frac{|f(x) - f(y)|}{|x-y|} \to 0$ as $x \to y$, implying that $f'(x) = 0$ everywhere in $(a, b)$, so $f$ is constant.

Would you like further details or have any questions about this proof? Here are some related questions to explore:

1. What is the role of the constant $C$ in this inequality?
2. How does this argument depend on differentiability of $f$?
3. Could this argument work if the inequality used $|x-y|$ instead of $|x-y|^2$?
4. Can $f$ have discontinuities under these conditions?
5. How does the Mean Value Theorem support this conclusion?

Tip: For functional inequalities, analyzing the behavior of the difference quotient is often key to unlocking proofs about derivatives.