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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?

jerry guna
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  • This inequality is certainly not true. As $x \to 0^+$ the left hand inequality diverges to $+\infty$ but the right hand inequality converges to 1. – JackT Sep 17 '21 at 12:35
  • Oops I edited now – jerry guna Sep 17 '21 at 12:44
  • Are you sure you want $\Gamma (x) \leq 1+1/x$? Note that even the harder ineq. $\Gamma (x) \leq 1/x$ is true, see here: https://math.stackexchange.com/questions/2205771/gamma-function-inequality-gammax-le-frac1x?rq=1 – Andreas Sep 17 '21 at 13:48
  • Yes that's what given in book. – jerry guna Sep 17 '21 at 13:54

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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 $\approx 0.8856$.)

Martin R
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