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?
 A: I think I have found a proof. Any comments, remarks are welcome.
We use the notation $x\wedge y=\min(x,y).$ Let $f:\mathbb{C}^2\to\mathbb{R}$ be the function defined by
$$
f(\varepsilon,\delta) = \left\{
    \begin{array}{ll}
        \frac{|\delta|}{\sqrt{|\varepsilon|}}\wedge \sqrt{|\delta|} & \mbox{when } \varepsilon\neq0, \\
        \sqrt{|\delta|} & \mbox{otherwise.}
    \end{array}
\right.
$$
$f$ is continuous since $\lim\limits_{\varepsilon\to 0}\frac{|\delta|}{\sqrt{|\varepsilon|}}\wedge \sqrt{|\delta|}= \sqrt{|\delta|}$ and the $\min,|\cdot|,\sqrt{\cdot}$ are continuous functions. Considering $f$ instead of the original RHS allows us to avoid undetermined forms.
Claim: For all $\varepsilon,\delta\in\mathbb{C}$ and any complex square root $\sqrt{\cdot},$
$$|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|\wedge|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|\leq f(\varepsilon,\delta).$$
Proof:
Throughout the proof, we use the fact that for any $z\in\mathbb{C},$ $|\sqrt{z}|=\sqrt{|z|}.$
Step 1)  Suppose $\varepsilon=0.$ Then the inequality becomes
$$\sqrt{|\delta|}\leq\sqrt{|\delta|},$$
which is satisfied for any $\delta\in\mathbb{C}.$
Step 2) Suppose $\varepsilon\neq 0.$ For all non-negative reals $a,b\geq 0,$
$$\sqrt{ab}\geq a\wedge b,$$
so that
$$|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|\wedge|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|\leq\sqrt{|\varepsilon+\delta-\varepsilon|}=\sqrt{|\delta}|.$$
This yields the first part of the inequality.
Note that
$$2\sqrt{|\varepsilon|}=|\sqrt{\varepsilon}-\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}+\sqrt{\varepsilon+\delta}|\leq|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|+|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|$$
by the triangle inequality. Suppose now 
$$|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|+|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|=0.$$
In turn, this implies
$$\varepsilon =0,$$
which contradicts our assumptions. So suppose
$$|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|+|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|>0.$$
Now for real numbers $a,b\geq 0$ such that $a+b\neq 0,$
$$\frac{2ab}{a+b}\geq\frac{ab}{a\vee b}=a\wedge b.$$
This finally yields
$$|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|\wedge|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|\leq \frac{2|\delta|}{|\sqrt{\varepsilon+\delta}-\sqrt{\varepsilon}|+|\sqrt{\varepsilon+\delta}+\sqrt{\varepsilon}|}\leq\frac{|\delta|}{\sqrt{|\varepsilon|}},$$
and concludes the proof.
