Do we have $\log\Gamma(2x+1)-2\log\Gamma(x+1)\ge\log(x^2+1)$ for $x\in[0,1]$?

Question

Prove or disprove that, for all $x \in [0, 1]$, $$\log\Gamma(2x+1)-2\log\Gamma(x+1)\ge\log(x^2+1).$$ Here $\Gamma$ is the gamma function.

I discovered this relation by accidentally drawing the graph of the two functions. Here is the image.

Given that the two functions intersect at both $x=0$ and $x=1$, is there any easy way to prove the inequality?

Answer 1

Proof.

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

Using (1), we have $$\ln \Gamma(2x + 1) = \int_0^\infty \frac{1}{t}\left(2x\mathrm{e}^{-t} + \frac{\mathrm{e}^{-t(2x+1)} - \mathrm{e}^{-t}}{1-\mathrm{e}^{-t}}\right)\,\mathrm{d}t, \tag{2}$$ and $$\ln\Gamma(x+1) = \int_0^\infty \frac{1}{t}\left(x\mathrm{e}^{-t} + \frac{\mathrm{e}^{-t(x+1)} - \mathrm{e}^{-t}}{1-\mathrm{e}^{-t}}\right)\,\mathrm{d}t. \tag{3}$$

Using the known identity (for $u > 0$) $$\ln u = \int_0^\infty \frac{\mathrm{e}^{-t} - \mathrm{e}^{-ut}}{t}\,\mathrm{d} t,$$ we have $$\ln(x^2 + 1) = \int_0^\infty \frac{\mathrm{e}^{-t} - \mathrm{e}^{-(x^2 + 1)t}}{t}\,\mathrm{d} t. \tag{4}$$

Using (2), (3) and (4), we have \begin{align*} &\ln \Gamma(2x+1)-2\ln\Gamma(x+1) - \ln(x^2+1)\\[6pt] ={}& \int_0^\infty \frac{\mathrm{e}^{-t} + \mathrm{e}^{-2tx} - 2\mathrm{e}^{-tx} + \mathrm{e}^{-tx^2} - \mathrm{e}^{-t(x^2 + 1)}}{t(\mathrm{e}^t - 1)}\,\mathrm{d} t. \tag{5} \end{align*}

From (5), it suffices to prove that, for all $t \ge 0$ and $x\in [0, 1]$, $$\mathrm{e}^{-t} + \mathrm{e}^{-2tx} - 2\mathrm{e}^{-tx} + \mathrm{e}^{-tx^2} - \mathrm{e}^{-t(x^2 + 1)} \ge 0. \tag{6}$$

We have $$\mathrm{e}^{-tx^2} - \mathrm{e}^{-t(x^2 + 1)} = \mathrm{e}^{tx(1 - x)}\mathrm{e}^{-tx}(1 - \mathrm{e}^{-t}) \ge [1 + tx(1 - x)]\mathrm{e}^{-tx}(1 - \mathrm{e}^{-t}) \tag{7}$$ where we use $\mathrm{e}^{u} \ge 1 + u$ for all $u \in \mathbb{R}$.

From (6) and (7), it suffices to prove that $$\mathrm{e}^{-t} + \mathrm{e}^{-2tx} - 2\mathrm{e}^{-tx} + [1 + tx(1 - x)]\mathrm{e}^{-tx}(1 - \mathrm{e}^{-t}) \ge 0$$ or (multiplying both sides by $\mathrm{e}^{tx}$) $$\mathrm{e}^{-t(1 - x)} + \mathrm{e}^{-tx} - 2 + [1 + tx(1 - x)](1 - \mathrm{e}^{-t}) \ge 0. \tag{8}$$

Clearly, we only need to prove the case that $x\in [0, 1/2]$.

Let $$f(x) := \mathrm{e}^{-t(1 - x)} + \mathrm{e}^{-tx} - 2 + [1 + tx(1 - x)](1 - \mathrm{e}^{-t}).$$ We have $$f'(x) = t\mathrm{e}^{-t(1 - x)} - t\mathrm{e}^{-tx} + t(1 - 2x)(1 - \mathrm{e}^{-t})$$ and $$f'''(x) = t^3(\mathrm{e}^{-t(1 - x)} - \mathrm{e}^{-tx}) \le 0.$$ Thus, $f'(x)$ is concave on $[0, 1/2]$. Also, we have $f'(0) = f'(1/2) = 0$. Thus, we have $f'(x) \ge 0$ on $[0, 1/2]$. Also, $f(0) = 0$. Thus, $f(x) \ge 0$ on $[0, 1/2]$.

We are done.