Let $s>0$ be a fixed real number.
Does $$ \inf_{x,y>0,\, xy \ge \frac{1}{4},\,x-y=s}(\sqrt{x}-\sqrt{y})^2\big( (\sqrt{x}+\sqrt{y})^2-2\big)=0\,\,\,?$$
Note that by the AM-GM inequality, we have $(\sqrt{x}+\sqrt{y})^2-2 \ge 0$, since $xy \ge \frac{1}{4}$.
In fact $(\sqrt{x}-\sqrt{y})^2\big( (\sqrt{x}+\sqrt{y})^2-2\big)>0$ since we assumed $x=y+s>y$.
I know that $\lim_{y \to \infty}(\sqrt{y+s}-\sqrt{y})=0$, but this does not seem to help immediately, since taking $y$ to be large makes $(\sqrt{x}+\sqrt{y})^2-2$ large as well.
$$ \left(\sqrt{x}-\sqrt{y}\right)^2\left(\left(\sqrt x+\sqrt y\right)^2-2\right) = (x-y)^2-2\left(\sqrt{x}-\sqrt{y}\right)^2 $$ Substituting $x=y+s$, this is equal to $$f(y)=s^2-2\left(\sqrt{y+s}-\sqrt{y}\right)^2=s^2-\frac{2s^2}{\left(\sqrt{y+s}+\sqrt{y}\right)^2}$$ and the domain is $y(y+s)\geq \frac 14, y>0\Rightarrow y\geq \frac{-s+\sqrt{s^2+1}}{2} $. It's not hard to see that $f$ is strictly increasing. So the infimum and supremum under these conditions are, respectively, $$f\left(\frac{-s+\sqrt{s^2+1}}{2}\right) $$ and $$\lim_{y\to\infty}f(y)=s^2 $$ The infimum will be a positive number because, as you said, the expression is always positive under those conditions.