Gamma function inequality

Question

I'm new to the gamma function and my current knowledge is that it's defined for all $z$ with $Re(z)>0$ by the integral

$$\Gamma(z)=\int_{0}^{\infty}t^{z-1}e^{-t}dt.$$

Recently, my little brother discovered the AM-GM inequality

$$\frac{x_1+x_2+...+x_n}{n}\geq (x_1x_2...x_n)^{\frac{1}{n}}.$$

If $x_1 = 1, x_2 = 2, ..., x_n = n$, the inequality can be rewritten to

$$n! \leq \left(\frac{n+1}{2}\right)^{n}.$$

For curiosity, I've decided to "extend" this inequality (maybe it's a famous one and I still don't know it). Here's my extension:

$$\Gamma(x+1) \leq \left(\frac{x+1}{2}\right)^{x}, \{x\in \mathbb{R}|x>-1\}$$

Using the site desmos.com, it's clear that the inequality is not valid at the interval $[0,1]$ (as can be seen in the figure).

So I think that the following is true:

$$ \boxed{\begin{array}_\Gamma(x+1)\leq \left(\frac{x+1}{2}\right)^{x}, \{x\in \mathbb{R}|-1<x\leq 0\,\cup\,x\geq1\}\\ \Gamma(x+1)\geq \left(\frac{x+1}{2}\right)^{x}, \{x\in \mathbb{R}|0\leq x\leq 1\}\end{array}} $$

I'm not good with inequalities, so I've decided to ask here: Can I prove the statement (if it's really true)? If so, how?

Thank you in advance. Really appreciate your attention.

Answer 1

$$\boxed{\begin{array}_\Gamma(x+1)\leq \left(\frac{x+1}{2}\right)^{x}, \{x\in \mathbb{R}|-1<x\leq 0\,\cup\,x\geq1\}\\ \Gamma(x+1)\geq \left(\frac{x+1}{2}\right)^{x}, \{x\in \mathbb{R}|0\leq x\leq 1\}\end{array}}$$

Proof:

We use the integral representation of $\ln\Gamma(x+1)$: $$\ln\Gamma(x+1) = \int_0^\infty \frac{1}{t}\left(x\mathrm{e}^{-t} + \frac{\mathrm{e}^{-t(x+1)} - \mathrm{e}^{-t}}{1-\mathrm{e}^{-t}}\right)\,\mathrm{d}t. \tag{1}$$ (See: https://functions.wolfram.com/GammaBetaErf/LogGamma/07/01/01/.)

Using the known identity $$\ln u = \int_0^\infty \frac{\mathrm{e}^{-t} - \mathrm{e}^{-ut}}{t}\,\mathrm{d} t,$$ we have $$x\ln\frac{x+1}{2} = \int_0^\infty x\cdot \frac{\mathrm{e}^{-t} - \mathrm{e}^{-(x+1)t/2}}{t}\,\mathrm{d} t. \tag{2}$$

Using (1) and (2), we have \begin{align*} &\ln\Gamma(x + 1)- x\ln \frac{x+1}{2}\\[6pt] =\,& \int_0^\infty \frac{1}{t}\left(\frac{\mathrm{e}^{-t(x+1)} - \mathrm{e}^{-t}}{1-\mathrm{e}^{-t}} + x\mathrm{e}^{-(x+1)t/2}\right)\,\mathrm{d} t\\[6pt] =\,&\int_0^\infty x\cdot \frac{\mathrm{e}^{-t}\mathrm{e}^{-xt/2}}{1-\mathrm{e}^{-t}}\left( \mathrm{e}^{t/2}\cdot \frac{1-\mathrm{e}^{-t}}{t} - \mathrm{e}^{xt/2}\cdot \frac{1-\mathrm{e}^{-xt}}{xt}\right)\,\mathrm{d} t. \tag{3} \end{align*}

Let $$f(u) := \mathrm{e}^{u/2}\cdot \frac{1-\mathrm{e}^{-u}}{u}.$$ It is easy to prove that $f(u)$ is strictly decreasing on $[-1, 0)$, and strictly increasing on $(0, \infty)$. Thus, we have $$\mathrm{e}^{t/2}\cdot \frac{1-\mathrm{e}^{-t}}{t} - \mathrm{e}^{xt/2}\cdot \frac{1-\mathrm{e}^{-xt}}{xt} \ge 0, \quad \forall t > 0, ~\forall x \in (-1, 0)\cup (0, 1]$$ and $$\mathrm{e}^{t/2}\cdot \frac{1-\mathrm{e}^{-t}}{t} - \mathrm{e}^{xt/2}\cdot \frac{1-\mathrm{e}^{-xt}}{xt} \le 0, \quad \forall t > 0, ~ \forall x \ge 1.$$

From (3), the desired results follow.

We are done.

Answer 2

If we consider the function $$f(x)=\Gamma(x+1) - \left(\frac{x+1}{2}\right)^{x}$$ we have

• $f(0)=0$ $$f'(0)=\log (2)-\gamma>0\qquad f''(0)=-2+\gamma ^2+\frac{\pi ^2}{6}-\log ^2(2)<0$$
• $f(1)=0$ $$ f'(1)=\frac{1}{2}-\gamma<0\qquad f''(1)=-1-2 \gamma +\gamma ^2+\frac{\pi ^2}{6}<0$$

All of the above make that there is a maximum somewhere between $0$ and $1$ and it is positive. This make your conjecture perfectly correct.

Edit

If we approximate $$f(x)=x(1-x)(a x+b)$$ and adjust $(a,b)$ in order to match the derivatives at $x=0$ and $x=1$, we have $$a=-\frac{1}{2}+2 \gamma -\log (2) \qquad\qquad b=\log (2)-\gamma$$

The maximum of $g(x)$ accurs at $x_*=0.4513$ which is not bad since the maximum of $f(x)$ accurs at $x_*=0.4479$.

Similarly, the maximum value of $g(x)$ is $0.0244$ while the maximum value of $f(x)$ is $0.0204$