Working a bit on
About the inequality conjectured as $x!>\left(\arctan\left(\cosh\left(x\right)\right)\right)^{a}$ for $x>0$ and fixed $a$
I got the inequality:


My first attempt was to translate into an integral reprentation as we have:

$${\displaystyle \int _{0 }^{\infty }e^{-x^{2}}\,dx=\frac{1}{2}{\sqrt {\pi }}.}$$


$${\displaystyle \Gamma (z)=\int _{0}^{\infty }x^{z-1}e^{-x}dx}$$

and compare the different integrals.

Some other information :

We have a quite good approximation of the minimum of the Gamma function taking :

$$(\pi-e+0.5)/2\simeq x_{min}=0.4616\cdots$$

Ps: $\Gamma(0.5)=\frac{\sqrt{\pi}}{2}$

Second attempt :

As it's hard to find the right trick to compare the two integrals , I have the idea to express the inequality as :


And use this link where we have :

$$ \Gamma(p)\Gamma(q)=4\int_0^{\infty}\int_0^{\infty}x^{2p-1}y^{2q-1}\operatorname{e}^{-x^2-y^2}\operatorname{d}\!x\operatorname{d}\!y.\tag 1 $$

And use :


Unfortunetaly I cannot procedd further .

Last attempt :

It seems we have the inequality on $[0,\pi-e]$: $$f(x)=\frac{\left(\left(\frac{x^{2}+1}{x+1}\right)^{45.9}+\left(\frac{x^{2}+2}{x+2}\right)^{45.9}\right)^{\frac{1}{45.9}}}{\left(2\right)^{\frac{1}{45.9}}}\leq x!$$

Remains to show :


Wich is (numerically) true.

I would like to know three things:

  1. How to finish it by hand ?
  1. Have you an alternative proof ?
  1. Is it a well-know result ?

Thanks in advance.

  • 4
    $\begingroup$ The absolute error is about $10^{-5}$. A proof will therefore be quite difficult. $\endgroup$
    – Peter
    May 30, 2022 at 12:13
  • $\begingroup$ $\Gamma(3/2)$ and $\Gamma(\pi -e + 1)$ are both very close to the minimum (within 1.8%) and equidistant from the minimum (approx. 0.12% difference). Since $\psi^{(2)}(x_{min}) < 0$, the effect of the third derivative should dominate and make $\Gamma(3/2) > \Gamma(\pi - e + 1)$. Not sure how to formalize this. $\endgroup$ Jun 2, 2022 at 17:34
  • $\begingroup$ @eyeballfrog Thanks for your comment I have the same point of view as you since I use strong convexity (wich is unsufficient) .At the end isn't it the opposite inequality ? $\endgroup$
    – DesmosTutu
    Jun 3, 2022 at 9:20

2 Answers 2


Suppose that you consider the function $$f(x)=(x-e)!-\frac {1} 2\sqrt{x}$$ and you search for its zero.

Let $x=y+e$ and consider that you look for the zero of function $$g(y)=\Gamma(y+1)-\frac 12 \sqrt{y+e}$$ $$g'(y)=\Gamma (y+1)\, \psi^{(0)} (y+1)-\frac{1}{4 \sqrt{y+e}}$$ $$g''(y)=\frac{1}{8 (y+e)^{3/2}}+\Gamma (y+1) \left(\psi ^{(0)}(y+1)^2+\psi ^{(1)}(y+1)\right)$$

By inspection $$g'(0)=-\frac{1}{4 \sqrt{e}}-\gamma$$ $$g'\left(\frac{1}{2}\right)=\frac{\sqrt{\pi }}{2} (2-\gamma -2 \log (2))-\frac{1}{4 \sqrt{\frac{1}{2}+e}}\quad <0$$ $$g'(1)=1-\frac{1}{4 \sqrt{1+e}}-\gamma\quad >0$$

$$\left|\frac{g'(1)}{g'\left(\frac{1}{2}\right)}\right| \sim 2.74$$ So the derivative cancels closer to $\frac{1}{2}$ than to $1$.

For $y_0=\frac 12$, all derivatives involves known numbers; so, we can use one single iteration of Newton-like methods of order $n$ and have explicit formulae for any $y_{(n)}$.

Adding $e$ to the result and computing their decimal representation, the results are $$\left( \begin{array}{ccc} n & x_{(n)} & \text{method} \\ 2 & 3.117817 & \text{Newton} \\ 3 & 3.146499 & \text{Halley} \\ 4 & 3.140573 & \text{Householeder} \\ 5 & 3.141912 & \text{no name} \\ 6 & 3.141614 & \text{no name} \\ \cdots & \cdots & \\ \infty & 3.141668 \end{array} \right)$$

  • 5
    $\begingroup$ Is this a proof of the inequality or a numeric calculation which makes it plausible/likely that the inequality holds? $\endgroup$
    – Martin R
    May 30, 2022 at 14:30
  • 1
    $\begingroup$ @MartinR. You can, for sure, consider that it is not a proof. However, there is absolutely no numerical calculations. Since we know all derivatives at $y_0=\frac 12$, $y_1$ is strictly analytic (except for $n \to \infty$). I could hev done much better assuming that we know the value of some $\Gamma$ for simple rational arguments. $\endgroup$ May 30, 2022 at 14:41

Too long for a comment :

It seems we have the inequality on $[0,1]$ :

$$f\left(x\right)=\frac{\left(\left(\frac{\left(x^{4}+b\right)}{x^{3}+b}\right)^{a}+\left(\frac{\left(x^{2}+0.75\right)}{x+0.75}\right)^{a}\right)^{\frac{1}{a}}}{2^{\frac{1}{a}}}\leq x!$$

Where $a=1.1605$ and $b=2$

Remains to show the inequality :


As we have removed the factorial it seems easier but I'm not sure because this last difference is circa $8.6*10^{-8}$ so we need strong inequalities .

Some other path :

Using strong convexity (and modifying it a bit) of $\Gamma(x+1)$ around $x=0.5$ we have :


Where :


Here I used this link Prove that $\ln2<\frac{1}{\sqrt[3]3}$ and the comparison with the Euler-Mascheroni constant and $\frac{1}{\sqrt{3}}$

Hope someone can achieve this .

  • $\begingroup$ We have the inequality : $$(\pi-e)!>g(\pi-e)>\sqrt{\pi}/2$$ where $$g\left(x\right)=\frac{\sqrt{\pi}}{2}\cdot\left(-2\ln\left(2\right)-0.5772+2\right)\left(x-0.5\right)+\frac{\sqrt{\pi}}{2}+\frac{\sqrt{\pi}}{4}\left(\left(2\ln\left(2\right)+\frac{1}{\sqrt{3}}-2\right)^{2}-4+\frac{\pi^{2}}{2}\right)\left(x-0.5\right)^{2}+\frac{f'''\left(0.5\right)}{6}\left(x-0.5\right)^{3}$$ and $f(x)=x!$ $\endgroup$
    – DesmosTutu
    Jun 3, 2022 at 9:37

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