<s>[INST] A simple inequality, does it always hold? I am looking at the following inequality:
$$\Gamma(-\frac{x+1}{x})\lt x , \forall x\gt \frac{7}{2}$$
It seems that the LHS and RHS eventually diverge for large enough $x$, but I have failed in a proof of that inequality. Any help would be greatly appreciated. Thanks.
 [/INST] We shall show this using the helpful observation made by Claude Leibovici.
First, we shall show that $0 < z < 2/7$,
$$
f(z) \equiv \log\Gamma(1 - z) - z + \frac{ z^2 } { 2} < 0.
\qquad (1)
$$
From the Bohr-Mollerup theorem, $\log\Gamma(1-z)$ is convex, so is $-z + z^2/2$.
So $f(z)$ is convex, and for $z < a = 2/7$, we have
$$
f(z) < \frac{z}{a} f(a) + \left(1 - \frac{z}{a}\right) f(0) = \frac{z}{a} f(a) < 0,
$$
where we have used the fact $f(a) \approx -0.0012 < 0$.
Second, for $0 < z < 1$
$$
\begin{aligned}
\log(1 + z)
&= z - \frac{z^2}2 + \frac{z^3}{3} - \frac{z^4}{4} + \dots
\\
&> z - \frac{z^2}{2} + \frac{z^3(1-z)}{4} + \frac{z^5(1-z)}{6} \dots
\\
&> z - \frac{z^2}{2}.
\qquad (2)
\end{aligned}
$$
Thus, from (1) and (2) we have for $0 < z < 2/7$,
$$
\log \frac{\Gamma(1-z)}{1+z} < f(z) < 0,
$$
or
$$
\frac{\Gamma(1-z)}{1+z} < 1.
$$
Finally, for $x > 7/2$,
$$
\begin{aligned}
\Gamma\left(-\frac{x+1}{x}\right)
&=
\frac{ \Gamma\left( -\frac 1 x \right) }
{\left(-1 - \frac 1 x \right)}
=
\frac{ \Gamma\left( 1 -\frac 1 x \right) }
{\left(-1 - \frac 1 x \right)\left(-\frac 1 x\right) }
=
x \, \frac{ \Gamma\left( 1 -\frac 1 x \right) } {1 + \frac{1}{x} }
< x.
\end{aligned}
$$
</s>