Claim :

Let $0.5-\varepsilon_k\leq x\leq 0.5+\varepsilon_k$ and $0<\varepsilon_k\leq 0.25$ with $k\geq 2$ then prove or disprove that :

$$\frac{\left(2^{-\left(2\left(1-x\right)\right)^{k}}\cdot x\right)}{1-2^{\left(k-1\right)\ln\left(2\right)}\left(\left(1-x\right)\cdot2\cdot x\right)^{\left(\left(k-1\right)\ln\left(2\right)+1\right)}}\geq x^{\left(2\left(1-x\right)\right)^{k}}$$

Background :

Again and I can say it's my last attempt on it I try to show :

Let $0<x<1$ and $k\geq 1$ then we have :

$$ x^{\left(2\left(1-x\right)\right)^{k}}+(1-x)^{\left(2\left(x\right)\right)^{k}}\leq 1$$

To show it we can directly use the Bernoulli's inequality on $0.5+\beta_k\leq x\leq1$ and use the claim if proved on $0.5-\varepsilon_k\leq x\leq 0.5+\varepsilon_k$ so the two inequalities complements each other .And the last inequality resulting from the claim is really easier than the Cirtoaje's inequality .

To finish I have tried to show the claim using logarithm and derivatives without reach the goal

Case $k=2$ :

As first step I simply use convexity because we have the inequality on $x\in[0.5,0.55]$:

$$\left(\frac{\left(r\left(0.5\right)-r\left(0.55\right)\right)}{-0.05}\left(x-0.5\right)+r\left(0.5\right)\right)^{2}\cdot x\leq \left(2^{-\left(2\left(1-x\right)\right)^{2}}\cdot x\right)$$

Where :


Then it seems we have on $x\in[0.54,0.55]$:

$$\left(\frac{\left(r\left(0.5\right)-r\left(0.55\right)\right)}{-0.05}\left(x-0.5\right)+r\left(0.5\right)\right)^{2}\geq \left(1-2^{\left(2-1\right)\ln\left(2\right)}\left(\left(1-x\right)\cdot2\cdot x\right)^{\left(\left(2-1\right)\ln\left(2\right)+1\right)}\right)\cdot x^{\left(4\left(1-x\right)^{2}\right)-1}$$

But It's really not convincing...because the derivatives are not equal at $x=0.5$

A better way would be to re-write the inequality as :

$$f\left(x\right)=\frac{0.5}{1-2^{\left(2-1\right)\ln\left(2\right)}\left(\left(1-x\right)\cdot2\cdot x\right)^{\left(\left(2-1\right)\ln\left(2\right)+1\right)}}\geq g(x)=\left(\left(2x\right)^{\left(\left(2\left(1-x\right)\right)^{2}-1\right)}\right)$$

Then take the logarithm ,make the difference and see what happens with :


It seems that the derivative of the difference admits three roots whose at $x=0.5$ .It seems it's also the case in general .

We can also substitute $y=2x$.

Last edit in this case ($k=2$) :

Using concavity and chord we have $x\in[0.5,0.55]$:

$\left(\frac{\left(f\left(0.5\right)-f\left(0.55\right)\right)}{-0.05}\left(x-0.5\right)+f\left(0.5\right)\right)^{\frac{1}{a\ln^{2}\left(1-x\right)}}\leq \frac{1}{1-2^{\left(2-1\right)\ln\left(2\right)}\left(\left(1-x\right)\cdot2\cdot x\right)^{\left(\left(2-1\right)\ln\left(2\right)+1\right)}}\quad (I)$

Where :

$$f\left(x\right)=\left(\frac{1}{1-2^{\left(2-1\right)\ln\left(2\right)}\left(\left(1-x\right)\cdot2\cdot x\right)^{\left(\left(2-1\right)\ln\left(2\right)+1\right)}}\right)^{a\ln^{2}\left(1-x\right)}$$

And $a$ evaluate as follow let:



Then :


Now it seems we have $x\in[0.5,0.55]$:

$$\left(\frac{\left(f\left(0.5\right)-f\left(0.55\right)\right)}{-0.05}\left(x-0.5\right)+f\left(0.5\right)\right)^{\frac{1}{a\ln^{2}\left(1-x\right)}}\cdot0.5\geq \left(2x\right)^{\left(\left(2\left(1-x\right)\right)^{2}-1\right)}$$

Ps : I change the last edit because before it was useless and not interesting now it's better .

PPS: The inequality $(I)$ must be restricted on $x\in[0.5,0.515]$ where the function $f(x)$ seems to be concave .

Question :

How to prove or disprove the claim ? How to determine $\varepsilon_k,\beta_k$ with some accuracy ?

Thanks for your effort in this sense .

  • 5
    $\begingroup$ “Last attempt“ – do you promise? I'll bookmark this and remind you :) $\endgroup$
    – Martin R
    Sep 6, 2021 at 12:15
  • $\begingroup$ OK. Now back to symmetry ! Cheers :-) $\endgroup$ Sep 6, 2021 at 13:27
  • 2
    $\begingroup$ @MartinR lol. By the way, the author does not change the name Erik Satie for a long time. It is nice. I noted them down: The.old.crap, The.old.boy, Mister.Expandead, Biggus Dickus Python, user698573, c-love-garlic, Erik Satie (Maybe FatsWallers, I am not sure). $\endgroup$
    – River Li
    Sep 6, 2021 at 13:27
  • 1
    $\begingroup$ @MartinR No worries :-)! For the news I shall open a bounty if you have somes ideas ... $\endgroup$ Sep 7, 2021 at 17:52
  • $\begingroup$ Using Gauss's curve we have : $m\left(x\right)=\frac{0.5}{1-2^{\left(2-1\right)\ln\left(2\right)}\left(\left(1-x\right)\cdot2\cdot x\right)^{\left(\left(2-1\right)\ln\left(2\right)+1\right)}}\geq g\left(x\right)=f\left(x\right)-f\left(0.5\right)+1$ where $f\left(x\right)=0.25+\frac{1}{b\cdot\sqrt{2\cdot\pi}}e^{-\frac{1}{2}\left(\frac{\left(x-a\right)}{b}\right)^{2}}$ and $a=0.5$ and $b=0.3$ $\endgroup$ Sep 9, 2021 at 18:01

2 Answers 2


Easy to see, difficult to "prove". Pictorial comment. Please don't vote.

enter image description here

$$ \Large \color{green}{x^{\left(2\left(1-x\right)\right)^{k}}}+\color{red}{(1-x)^{\left(2\left(x\right)\right)^{k}}}\leq 1 $$

  • 1
    $\begingroup$ Many inequalities are “easy to see” but difficult to prove. I honestly do not see how this “pictorial comment” helps to prove the claim. $\endgroup$
    – Martin R
    Sep 16, 2021 at 8:26

I get something interesting in the general case let's go !

Claim :

Let $0.5\leq x\leq 0.5+\varepsilon_k$ and $0<\varepsilon_k\leq 0.05$ with $k\geq 2$

Then it seems we have :

$\left(\frac{\left(f\left(0.5\right)-f\left(0.5+\varepsilon_k\right)\right)}{-\varepsilon_k}\left(x-0.5\right)+f\left(0.5\right)\right)^{\frac{1}{a\ln^{2}\left(1-x\right)}}\leq \frac{1}{1-2^{\left(k-1\right)\ln\left(2\right)}\left(\left(1-x\right)\cdot2\cdot x\right)^{\left(\left(k-1\right)\ln\left(2\right)+1\right)}}\quad (I)$

Where :

$$f\left(x\right)=\left(\frac{1}{1-2^{\left(k-1\right)\ln\left(2\right)}\left(\left(1-x\right)\cdot2\cdot x\right)^{\left(\left(k-1\right)\ln\left(2\right)+1\right)}}\right)^{a\ln^{2}\left(1-x\right)}$$

And $a$ evaluate as follow let:



Then :


Now it seems we have $x\in[0.5,0.5+\varepsilon_k]$:

$$\left(\frac{\left(f\left(0.5\right)-f\left(0.5+\varepsilon_k\right)\right)}{-\varepsilon_k}\left(x-0.5\right)+f\left(0.5\right)\right)^{\frac{1}{a\ln^{2}\left(1-x\right)}}\cdot0.5\geq \left(2x\right)^{\left(\left(2\left(1-x\right)\right)^{k}-1\right)}$$

This is a conjecture .The main idea is based on the fact that we use a derivative wich is also a chord because on $x\in[0.5,0.5+\varepsilon_k]$ the function $f(x)$ is neither concave nor convexe.In fact it seems there is just one minimum ($f''(x)=0$) on $x\in[0.5,0.5+\varepsilon_k]$ .

As Tyma Gaidash says correct me and give me feedback !

Edit :

Using Binomial inequality in the case $k=2$ we have :

Let $x\in[0.35,0.65]$ then we have :

$$1+\left(4\left(1-x\right)^{2}-1\right)\left(2x-1\right)+\left(-1+4\left(1-x\right)^{2}\right)\left(-1+4\left(1-x\right)^{2}-1\right)\left(2x-1\right)^{2}\cdot0.5\geq \left(2x\right)^{-1+4\left(1-x\right)^{2}}$$

Why the last inequality is interesting ? Because we have an equality case at $x=0.5$ wich is also true with Bernoulli's inequality but in this case Bernoulli's inequality is unsufficient and we need use a second order approximation. It's also interesting because it's not the case with Cirtoaje's inequality where the equality case in the Bernoulli's inequality is reached for $x=1$ .So I think it's a good trick .

Proof :

It's an direct application of the Gerber's inequality

Gerber's inequality :

If $\alpha\in R$ , and $n$ a natural number and $x>-1$ then :

$$\binom{\alpha}{n+1} x^{n+1}\leq 0\implies (1+x)^{\alpha}\leq \sum_{i=0}^{n}\binom{\alpha}{i}x^i$$

Edit 2 :

I go a little bit further it seems we have the following inequality on $x\in[0.5,0.58]$ :

$$k(x)=0.5\left(1+\left(x\left(1-x\right)\right)^{1+3.44\ln\left(2\right)}\cdot4^{1+3.44\ln\left(2\right)}\right)\leq \frac{0.5}{1-2^{\ln\left(2\right)}\left(2x\left(1-x\right)\right)^{1+\ln\left(2\right)}}$$

And $$k(x)\geq 1+\left(4\left(1-x\right)^{2}-1\right)\left(2x-1\right)+\left(-1+4\left(1-x\right)^{2}\right)\left(-1+4\left(1-x\right)^{2}-1\right)\left(2x-1\right)^{2}\cdot0.5=32(x-1)x((x-2)x+0.75)^2+1$$

Last edit : a clap for the end :

Again I go a little bit further .

We have the following inequality on $x\in(0,1)$ :

$$2^{-4\left(1-x\right)^{2}}\cdot x\leq\left(p\left(2\left(1-x\right)^{2}\right)\right)^{-1}\cdot x$$

Where :


Proof :

Use the Taylor series of $2^{2x}$ around $x=0.5$ and a bit of algebra .

Now it seems we have on $x\in(0.25,0.75)$ :

$$1-2^{\ln(2)}((1-x)2x)^{\ln(2)+1}-q\left(1-x\right)-q\left(x\right)\geq 0$$

Where :

$$q\left(x\right)=\left(p\left(2\left(1-x\right)^{2}\right)\right)^{-1}\cdot x$$


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