Proof of $\Gamma(x)\Gamma\left(\dfrac{1}{x}\right)\gt 1$

Question

Is it possible to prove the following inequality: $$\forall x\gt0,f(x)=\Gamma(x)\Gamma\left(\dfrac{1}{x}\right)\gt 1?$$ $f(x)$ has a minimum where $f(x)'=0$ which means: $$\Psi(x)\Gamma(x)\Gamma(1/x)=\dfrac{\Gamma(x)\Psi(1/x)\Gamma(1/z)}{x^2}$$ It seems quite impossible to solve this equation in a closed form. Assuming the solution is $x_m$, the minimum value of $f(x)$ shoud be $f(x_m)$ and if $f(x_m)$ is greater than one the inequality would have been proved. Is there another way to come at the same conclusion? Thanks.

Answer 1

If $$f(x)=\Gamma(x)\,\Gamma\left(\frac{1}{x}\right)$$ $$f'(x)=\frac{\Gamma \left(\frac{1}{x}\right) \Gamma (x) }{x^2}\left(x^2 \psi(x)-\psi \left(\frac{1}{x}\right)\right)$$ for which the only solution seems to be $x=1$.

Working the second derivative (I shall not write it), you would find that $$f''(1)=\frac{\pi ^2}{3}-2 \gamma >0$$ confirming that we face a minimum.

Answer 2

Here is another suggestion. I'm not sure if this would be "cleaner" than computing the derivative. For positive real $x$ you have:

$$ \Gamma(x)=\lim_{n\to \infty}\dfrac{n!n^x}{x(x+1)\ldots(x+n)}. $$

Since the limits exist for $x$ and $1/x$ you can multiply the terms together and try to make an inductive argument that the terms are bounded by $1$ from below.