# Gamma function inequality

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

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?

• I added two points in the edit of my answer. Commented Jan 8, 2023 at 3:34

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 $$0.0204$$

• Hi Claude. Thank you for this solution. It's great. I'm still wondering if there is an approach that does not require first and second derivatives (maybe just some inequalities). Commented Jan 8, 2023 at 18:02
• @otreblig I think this is not enough to prove that $f(x) \ge 0$ on $[0, 1]$. For example, consider $g(x) = x(1-x)(8x^2-9x+2)$. We have $g(0) = 0, g'(0) > 0, g''(0) < 0$ and $g(1) = 0, g'(1) < 0, g''(1) < 0$. However $g(1/2) < 0$. Commented Jan 9, 2023 at 2:48

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

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.

• Thank you! Really great answer. Can I ask you for a link or paper that contains the proof of the log gamma integral? Or it's just Binet's formula after some manipulation? Commented Jan 10, 2023 at 1:01
• @otreblig You may find it in the book "A Course of Modern Analysis" by E. T. Whittaker and G. N. Watson, Chapter "gamma function". Commented Jan 10, 2023 at 2:34