$\frac{\Gamma\left(x+\frac{1}{x}+1\right)}{2}\geq \left(\Gamma(x+1)\Gamma\left(\frac{1}{x}+1\right)\right)x^{-\frac{1}{2\pi}\ln\left(x\right)},x>0$

Question

It's a very challenging and accurate inequality let's say :

Let $x>0$ then we have :

$$\frac{\Gamma\left(x+\frac{1}{x}+1\right)}{2}\geq \left(\Gamma(x+1)\Gamma\left(\frac{1}{x}+1\right)\right)x^{-\frac{1}{2\pi}\ln\left(x\right)}$$

I attempt to use Stirling's approximation first but failed then .

I think we have more chance with the modified version of the Stirling's approximation due to Ramanujan .

Have you an idea to show or a draft ?

Edit :

Why it would be an interesting result ?

The question could be is here interesting pattern ?

The question is yes! there is something interesting

First $n<0$ a negative integer:

$$\frac{\left(n+\frac{1}{n}\right)!}{2n!\left(\frac{1}{n}\right)!}=0$$

It behave like $-\tan\left(\frac{x\pi}{2}\right)$ for negative value .

Moreover :

$\exists x$ such that :

$$\frac{\left(x+\frac{1}{x}\right)!}{2x!\left(\frac{1}{x}\right)!}=-\tan\left(\frac{x\pi}{2}\right)$$

Then the solutions $x$ seems to be on a line $\tan\left(12\right)x-0.475$.

Hint :

We have for $x>0$ :

$$\ln\left(\frac{\left(x+\frac{1}{x}\right)!}{2x!\left(\frac{1}{x}\right)!}\right)+1-x^{-\frac{1}{2\pi}\ln\left(x\right)}\le \left(\frac{\left(x+\frac{1}{x}\right)!}{2x!\left(\frac{1}{x}\right)!}\right)-x^{-\frac{1}{2\pi}\ln\left(x\right)}$$

Then as we have the Dixon's theorem :

$$\sum_{k}^{ }(-1)^{k}{a+b\choose a+k}{a+c\choose c+k}{b+c \choose c+k}=\frac{(a+b+c)!}{a!b!c!}$$

Or :

$$\sum_{k}^{ }(-1)^{k}{x+\frac{1}{x}\choose x+k}{x\choose k}{\frac{1}{x} \choose \frac{1}{x}+k}=\frac{(x+\frac{1}{x})!}{x!\left(\frac{1}{x}\right)!}$$

Setting $k=0$ other value gives zero we need to show $x>0$ : $$\ln\left({x+\frac{1}{x}\choose x}{x\choose 0}{\frac{1}{x} \choose \frac{1}{x}+0}\right)+1-x^{-\frac{1}{2\pi}\ln\left(x\right)}\geq 0$$

Then it seems we have for $x\in[1,1.25]$ :

$$(2-{1/x+x\choose 1}/2)^{\frac{1}{x\cdot \pi}}-x^{-\frac{1}{2\pi}\ln\left(x\right)}\leq {1/x+x\choose x}-2x^{-\frac{1}{2\pi}\ln\left(x\right)}$$

Then there is a possibility to use https://mathworld.wolfram.com/BinomialCoefficient.html at the end $m=1$

Using the mathsworld link it seems we have $x\in[1,1.25]$ :

$$\left(2-\frac{\left(x+\frac{1}{x}\right)^{\left(x+\frac{1}{x}\right)}}{4\left(x+\frac{1}{x}-1\right)^{-1+x+\frac{1}{x}}}\right)^{\frac{1}{x\cdot\pi\cdot\ln\left(4\right)}}-x^{-\frac{1}{2\pi}\ln\left(x\right)}\leq (2-{1/x+x\choose 1}/2)^{\frac{1}{x\cdot \pi}}-x^{-\frac{1}{2\pi}\ln\left(x\right)}$$

Answer 1

If $$f(x)=\text{lhs - rhs}\qquad \qquad f\left(\frac{1}{x}\right)=f(x)$$ the first derivative cancels at $x=1$ and $$f''(1)=3+\frac{1}{\pi }-\frac{\pi ^2}{3}\quad > \quad 0$$

Using a Taylor expansion around $x=1$

$$f(x)=\frac{\left(3+9 \pi -\pi ^3\right) }{6 \pi }\,(x-1)^2\,\Big(2-x+O\left((x-1)^2\right) \Big)$$

Edit

Close to $0$ the asymptotics is (more or less)

$$f(x) \sim \sqrt{\frac{\pi }{2x}} \exp\left(-\frac{\log (x)+1}{x} \right)$$

Edit Wolframandgromit to help ClaudeLeibovici :

Setting $0<x<1$ :

$$x^{-\frac{1}{2\pi}\ln\left(x\right)}\leq h\left(x\right)=\frac{1}{\frac{\ln^{2}\left(x\right)}{7}+1}\leq q(x)=f\left(x+\frac{1}{x}\right)-f\left(x\right)-f\left(\frac{1}{x}\right)+1-f\left(2\right)+2f\left(1\right),f\left(x\right)=\left(x+1\right)\ln\left(x+1\right)-\left(x+1\right)+\frac{1}{2}\ln\left(\frac{2\pi}{x+1}\right)$$

We introduce :

$$m\left(x\right)=g\left(x+\frac{1}{x}\right)-g\left(x\right)-g\left(\frac{1}{x}\right)+1-1-f\left(2\right)+2f\left(1\right)-\ln\left(2\right),g\left(x\right)=\int_{0}^{\infty}\frac{2\arctan\left(\frac{t}{x+1}\right)}{e^{2\pi t}-1}dt$$

$\exists x,p\in(0,1),\exists b>1$ such that :

$$r\left(x\right)=-m\left(1-bx\left(1-x\right)\right)+m\left(x\right)\ge p>0$$

Using my comment below

And :

$$k\left(x\right)=q\left(x\right)-q\left(1-bx\left(1-x\right)\right)-h\left(x\right)+h\left(1-b\left(1-x\right)x\right)\leq 0,b>1/x$$

Iterating gives the inequality .

Update

Close to $0$, a better asymptotic seems to be $$f(x)=\frac{1}{12} \sqrt{\frac{\pi }{2}} x^{-\frac{x+2}{2 x}} e^{-\frac{1}{x}-\frac{\log ^2(x)}{2 \pi }}\,A(x)$$ where $$A(x)=(24 \gamma -2) x+e^{\frac{\log ^2(x)}{2 \pi }} (x-12 x \log (x)+12)-24$$

Answer 2

$$\frac{(x+1/x)!}{2} \ge (x)!\cdot(1/x)!\cdot x^{\frac{-1}{2\pi}\log(x)}$$ say $(x)! = e^{f(x)}$ $$e^{f(x+1/x)}\cdot (1/2) \ge e^{(x)!}\cdot e^{(1/x)!}\cdot x^{\frac{-1}{2\pi}\log(x)}$$ $$f(x+1/x)+\log(1/2) \ge f(x)+f(1/x)+\log(x^{\frac{-1}{2\pi}\log(x)})$$ $$f(x+1/x)-\log(2) \ge f(x)+f(1/x)-\frac{1}{2\pi}\cdot \log(x)\cdot \log(x)$$ $$f(x+1/x) \ge f(x)+f(1/x)+\log(2)-\frac{1}{2\pi}\cdot \log(x)^2 $$ we know $f(x+1/x) \gt f(x)$ and because $f(x) = \log(x!)$, then $\color{red}{f(1/x)}$ would be negative value $$f(x+1/x) \ge f(x)+\log(2)+f(1/x)-\frac{1}{2\pi}\cdot \log(x)^2 $$ can you finish it up now

Answer 3

You can bring the given inequality in the form $$\frac{x^{c\ln x}}{2(x+\frac1x+1)}\geq B(x+1,\frac1x+1)$$ where $c>0$ or specially $c=\frac1{2\pi}$ and where $B(.,.)$ is the beta function.

It is not difficult to show that the $LHS$ has the minimum value $\frac16$ attained at $x=1$. So it is enough to show that $B(x+1,\frac1x+1)\leq\frac16$ for $x>0.$ I have not a proof yet but WolframAlpha confirms it.