Show the inequality $\frac{\sqrt{\pi}}{2}<\left(\pi-e\right)!$ 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:
$$\frac{\sqrt{\pi}}{2}<\left(\pi-e\right)!$$
My first attempt was to translate into an integral reprentation as we have:
$${\displaystyle \int _{0 }^{\infty }e^{-x^{2}}\,dx=\frac{1}{2}{\sqrt {\pi }}.}$$
and
$${\displaystyle \Gamma (z)=\int _{0}^{\infty }x^{z-1}e^{-x}dx}$$
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 :
$$\frac{\sqrt{\pi}}{2}\frac{\sqrt{\pi}}{2}<\frac{\sqrt{\pi}}{2}\left(\pi-e\right)!$$
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 :
$$\int_{0}^{\infty}\int_{0}^{\infty}e^{-\left(x^{2}+y^{2}\right)}dxdy=\frac{\pi}{4}$$
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 :
$$f(\pi-e)>\sqrt{\frac{\pi}{4}}$$
Wich is (numerically) true.
I would like to know three things:


*

*How to finish it by hand  ?





*Have you an alternative proof ?





*Is it a well-know result ?


Thanks in advance.
 A: 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
$$\left(
\begin{array}{ccc}
 n & x_{(n)}  & \text{method} \\
 2 &  3.117817 & \text{Newton} \\
 3 &  3.146499 & \text{Halley} \\
 4 &  3.140573 & \text{Householeder} \\
 5 &  3.141912 & \text{no name} \\
 6 &  3.141614 & \text{no name} \\
\cdots & \cdots & \\
\infty & 3.141668
\end{array}
\right)$$
A: 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 :
$$f\left(\pi-e\right)-\sqrt{\frac{\pi}{4}}>0$$
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 :
$$(\pi-e)!>h\left(\pi-e\right)>\frac{\sqrt{\pi}}{2}$$
Where :
$$h(x)=\frac{\sqrt{\pi}}{2}\cdot\left(-3\frac{1}{3^{\frac{1}{3}}}+\ln\left(2\right)-\frac{1}{\sqrt{3}}+2\right)\left(x-0.5\right)+\frac{\sqrt{\pi}}{2}+\frac{\sqrt{\pi}}{4}\left(\left(2\ln\left(2\right)+\frac{1}{\sqrt{3}}-2\right)^{2}-4+\frac{\pi^{2}}{2}\right)\left(x-0.5\right)^{2}$$
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 .
