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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?

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  • $\begingroup$ Are you looking for a demonstration only of the complex claim? $\endgroup$ – Brian Tung Jul 15 '16 at 18:11
  • $\begingroup$ Yes. I already have LHS $\leq \sqrt{|\delta|}.$ I will add details. $\endgroup$ – themightymoose Jul 15 '16 at 18:15
  • $\begingroup$ I just wrote the others as a follow-up but I have the proof. $\endgroup$ – themightymoose Jul 15 '16 at 18:33
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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.

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