radicals - Square root inequality revisited

This is a follow-up question of this one:
Proof of the square root inequality $2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}$

I am interested in the following generalizations of the square root inequality.
Let $\varepsilon,\delta>0.$ Then
$$\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}<\frac{\delta}{2\sqrt{\varepsilon}}.$$

If additionally $\delta<\varepsilon$, then
$$\frac{\delta}{2\sqrt{\varepsilon}}<\sqrt{\varepsilon}-\sqrt{\varepsilon-\delta}.$$ The proof is similar as answered in my previous question.

Moreover, if $\varepsilon,\delta\in\mathbb{C}$ are any complex numbers and $\sqrt{\cdot}$ is a complex root, then
$$\min(|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|,|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|)\leq\min(\frac{|\delta|}{\sqrt{|\varepsilon|}},\sqrt{|\delta|}).$$
I have trouble to prove this one, but it should hold (it is not from me). What I have yet is the following: first suppose
$$|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|\leq|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|.$$
Then
$$|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|^2\leq|\delta|,$$
so that

$$|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|\leq\sqrt{|\delta|}.$$
Similarly, when
$$|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|\geq|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|,$$
then
$$|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|\leq\sqrt{|\delta|},$$
so that
$$\min(|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|,|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|)\leq\sqrt{|\delta|}.$$
Moreover,
$$|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|=\frac{|\delta|}{|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|}\leq\ldots$$
and I don't know how to continue. Any help for solving this?