How to show that $(ex)^{-1}\leq \Gamma (x) \leq 1+1/x$, if $0<x<1$?

Question

We can use the fact that $\Gamma(x) =\frac{\Gamma (x+1) }{x}$. Suppose $e^{-1}\leq \Gamma (x) \leq x$, if $1<x<2$, then the required inequality can be proven. But is that inequality true? How to prove that?

Answer 1

An upper bound has been proven here: $\Gamma(x) < 1$ for $1 < x < 2$ follows from the convexity of the Gamma function, and therefore is $$ \Gamma(x) \le \frac 1 x \text{ for } 0 < x < 1 \, . $$

A simple lower bound can be obtained from the integral representation: For $1 < x < 2$ is $$ \Gamma(x) = \int_0^\infty t^{x-1} e^{-x} \, dx \ge \int_1^\infty t^{x-1} e^{-x} \, dx \\ \ge \int_1^\infty e^{-x} \, dx = \frac 1e $$ and therefore $$ \Gamma(x) \ge \frac 1 {ex} \text{ for } 0 < x < 1 \, . $$

(That is a very rough estimate. The minimal value of the Gamma function is $\approx 0.8856$.)