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Problem :

Show that :

$$\frac{1}{2\ln2}\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right)>\sqrt{2}-1$$



Using some approximation using itself algoritm found here (https://en.wikipedia.org/wiki/Methods_of_computing_square_roots)I have :

$$99/70-1>\sqrt{2}-1$$

Wich is too large .

On the other hand see this question for an approximation of $\ln 2$ Any good approximation methods of $\ln(2)$?



Update :

Using the inverse function (see my comment) it remains to show :

$$\ln2<\frac{1}{\sqrt{2}}$$

Update 2 :

Using my answer in this link (inequality due @MichaelRozenberg) Prove that $\ln2<\frac{1}{\sqrt[3]3}$ :

We have :

$$\ln2< \frac{1}{3^{\frac{1}{3}}}$$

Remains to show that :

$$\frac{1}{3^{\frac{1}{3}}}<\frac{1}{\sqrt{2}}$$

Wich is easy !

How to show by hand without the helps of a computer ?

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    $\begingroup$ Sorry but I have to ask again: Why would that be a useful result? $\endgroup$
    – Martin R
    Sep 18, 2022 at 9:47
  • $\begingroup$ @MartinR It could be a nice approximation of $\ln2$ using the inverse function see wolframalpha.com/… $\endgroup$ Sep 18, 2022 at 9:51
  • $\begingroup$ Unless I am mistaken, $\ln2<\frac{1}{\sqrt{2}}$ is equivalent to $\exp(\sqrt 2) > 4$, and that is easy to get from the Taylor series until the $x^4/4!$ term. $\endgroup$
    – Martin R
    Sep 18, 2022 at 10:42

2 Answers 2

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You may show the inequality using elementary equivalent rearrangements and Cauchy-Schwarz inequality as shown below.

This is surely not the nicest way but it works:

\begin{eqnarray*} \frac{1}{2\ln2}\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right) & > & \sqrt{2}-1 \\ & \Leftrightarrow & \\ \frac{1}{2\ln2}\left(1-\frac{1-\ln2}{1+\ln2}\right) & > & \left(\sqrt{2}-1\right)\left(1+\sqrt{\frac{1-\ln2}{1+\ln2}}\right) \\ & \stackrel{\frac 1{\sqrt 2 - 1} = \sqrt 2 + 1}{\Longleftrightarrow} & \\ \sqrt 2 + 1 & > & \left(1+\ln 2\right)\left(1+\sqrt{\frac{1-\ln2}{1+\ln2}}\right) \\ & \Leftrightarrow & \\ \sqrt 2 & > &\ln 2 + \sqrt{1-\ln^2 2} \\ \end{eqnarray*}

The last inequality is true because of Cauchy-Schwarz:

$$1\cdot \ln 2 + 1\cdot \sqrt{1-\ln^2 2} \stackrel{C.S.}{<}\sqrt 2 \cdot \sqrt{\ln^2 2 + 1 - \ln^2 2} = \sqrt 2$$

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A nice trick for your last question is this: take the exponential of both terms, that is

$$e^{\ln(2)} < e^{1/\sqrt{2}}$$ that is

$$2 < e^{1/\sqrt{2}}$$

Now if $x$ is small enough, you can use the $(2, 2)$ Padé approximant which has a really notable form:

$$e^x \approx \frac{(x+3)^2 + 3}{(x-3)^2+3}$$

Consider that if $|x| < 1/2$ (not your case but we're close, since $1/\sqrt{2} \approx 0.7$) the absolute difference between $e^x$ and Padé approximat is about $8\cdot 10^{-5}$.

In this case, plugging your value:

$$e^{1/\sqrt{2}} \approx \frac{(1/\sqrt{2}+3)^2 + 3}{(1/\sqrt{2}-3)^2+3} = \frac{1}{553} \left(300 \sqrt{2}+697\right) \approx 2.02$$

Which you really can calculate by pen and hand.

Hence $2 < 2.02$ as wanted.

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