Let $k>0$ be a constant chosen sufficiently large. In a problem I recently solved, I had to lower bound the function $$\mathcal S_k(x) := \sum_{n=0}^{\lfloor \sqrt{x} \rfloor} \left( \lfloor \sqrt{x+kx^{1/4}-n^2} \rfloor - \lfloor \sqrt{x-n^2} \rfloor \right)$$ by a constant. I plotted the same on Desmos and noticed it shows a general increasing trend as $x \rightarrow \infty$. So I got curious and tried to lower bound this by functions that go to infinity with $x$, but so far have obtained no such bound.
It seems like such a bound should not be too hard to obtain though. For example, if say we did not have the floors, then each term would be of the form $$\sqrt{x+kx^{1/4}-n^2} - \sqrt{x-n^2} = \frac{kx^{1/4}}{\sqrt{x+kx^{1/4}-n^2} + \sqrt{x-n^2}} \geq \frac{kx^{1/4}}{\sqrt{x+kx^{1/4}} + \sqrt{x}} \tag{1}$$ which would give me a lower bound of $$\frac{kx^{1/4} (\lfloor \sqrt{x} \rfloor + 1)}{\sqrt{x+kx^{1/4}} + \sqrt{x}} \gg x^{1/4}$$ on the (modified) $\mathcal S_k(x)$.
But with the floors, it becomes difficult to put a bound of the type $(1)$. I even tried to weaken the (analogous) sought estimate to something like $$\lfloor \sqrt{x+kx^{1/4}} \rfloor - \lfloor \sqrt{x} \rfloor \geq \frac{kx^{1/4-\epsilon}}{\lfloor \sqrt{x+kx^{1/4}} \rfloor + \lfloor \sqrt{x}\rfloor}$$ for some $\epsilon \in [0, 1/4)$ fixed, but then the inequality doesn't seem to be true anymore (at least not with $\epsilon \in [1/8, 1/4)$).
I also noticed that the sum $S_k(x)$ is basically the number of times for which $\sqrt{x-n^2}$ and $\sqrt{x+kx^{1/4}-n^2}$ does not lie between the same pair of consecutive integers, and tried to give some combinatorial bound based on that but that ventue hasn't gone anywhere as of yet.
I would really like to know if one of the above approaches would work, and more generally, if there is a function going to infinity with $x$ which lower bounds the sum $\mathcal S_k(x)$.