Prove the inequality $\ln\left(\Gamma(x^x+1)\right)< x^x\ln\left(\Gamma(x+1)\right)$ for $0<x<1$

Question

Following some toughts on Approximation of $\Big[\Gamma(1+x)\Big]^{-1}$ for $0 \leq x \leq 1$ (for the art for art's sake). .

I have the inequality :

Claim:

Let $0<x<1$ then we have :

$$\Gamma(x^x+1)< \Big(\Gamma(x+1)\Big)^{x^x}$$

The interesting part is near the value $x=1$.

First I rewrite the inequality as :

$$\ln\left(\Gamma(x^x+1)\right)< x^x\ln\left(\Gamma(x+1)\right)$$

Then I introduce the function :

$$f(x)=\ln\left(\Gamma(x^x+1)\right)-x^x\ln\left(\Gamma(x+1)\right)$$ and now I want to differentiate but it gives nothing good .I have tried also simple refinements without success.

My question :

How to prove the claim ?

Thanks in advance !

Answer 1

Not a final answer but too long for a comment.

Hoping that you do not mind, I shall define $$f(x)=x^x\log\left(\Gamma(x+1)\right)-\log\left(\Gamma(x^x+1)\right)$$

Close to $x=1$, we have $$f(x)=\left(\frac{\gamma }{2}+\frac{\pi ^2}{12}-1\right) (1-x)^3+$$ $$\frac{1}{12} (1-x)^4 \left(8-2 \gamma -\pi ^2-4 \psi ^{(2)}(2)\right)+O\left((1-x)^5\right)$$ which matches the function quite decently for $0.8 \leq x \leq 1.0$.

Close to $x=0$, we have $$f(x)=x ((\gamma -1) \log (x)-\gamma )+$$ $$\frac{1}{12} x^2 \left(\left(6 \gamma -\pi ^2\right) \log ^2(x)-12 \gamma \log (x)+\pi ^2\right)+O\left(x^3\right)$$which matches the function quite decently for $0.0 \leq x \leq 0.1$.

Using the first term only, this predicts a maximum value at $$x_*=e^{\frac{1}{\gamma -1}}\approx 0.0939237$$ and $f(x_*)=0.0330039$ while a rigorous optimization would give a maximum value of $0.0330818$ at $x=0.103099$.

Answer 2

Some thoughts:

Case 1 $~ 0 < x \le \frac15$:

It suffices to prove that $$\Gamma(x + 1) \ge \Gamma(x^x + 1).$$ Since $y \mapsto \Gamma(y + 1)$ is strictly increasing on $(1/2, 1]$, using $1/2 < x^x < 1$ for all $x\in (0, 1/5]$, noting also that $x^x \le 1 + \frac56 x\ln x$ for all $x\in (0, 1/5]$, it suffices to prove that $$\Gamma(x + 1) \ge \Gamma\left(2 + \frac56 x\ln x\right).$$

Fact 1: $\ln \Gamma(x + 1) \ge \frac{12x^3 - x^2 - 14x}{12x^2 + 36x + 24}$ for all $x > 0$.
(Hint: Take derivatives. Use (the digamma function) $\psi(u) \ge \ln u - \frac{1}{2u} - \frac{1}{12u^2}$ for all $u > 0$.)

Fact 2: For all $v\in [1, 2]$, $$\ln \Gamma(v) \le - \frac32\ln 2 + \frac{47}{24} + (v - 1/2)\ln v - v + \frac{1}{12v}.$$ (The proof is similar to that of Fact 1.)

Using Facts 1-2, it suffices to prove that \begin{align*} &\frac{12x^3 - x^2 - 14x}{12x^2 + 36x + 24}\\ \ge\, & - \frac32\ln 2 + \frac{47}{24} + \left(2 + \frac56 x\ln x - \frac12\right)\ln \left(2 + \frac56 x\ln x\right)\\ &\quad - \left(2 + \frac56 x\ln x\right) + \frac{1}{12(2 + \frac56 x\ln x)}. \end{align*}

Fact 3: For all $x \in (0, 1]$, $$\ln\left(2 + \frac56 x\ln x\right) \le \ln 2 + \frac{10x\ln x}{24 + 5x\ln x}.$$

Using Fact 3, it suffices to prove that \begin{align*} &\frac{12x^3 - x^2 - 14x}{12x^2 + 36x + 24}\\ \ge\, & - \frac32\ln 2 + \frac{47}{24} + \left(2 + \frac56 x\ln x - \frac12\right)\cdot \left(\ln 2 + \frac{10x\ln x}{24 + 5x\ln x}\right)\\ &\quad - \left(2 + \frac56 x\ln x\right) + \frac{1}{12(2 + \frac56 x\ln x)}\\ =\, & F(x\ln x) \end{align*} where $$F(u) = - \frac32\ln 2 + \frac{47}{24} + \left(2 + \frac56 u - \frac12\right)\cdot \left(\ln 2 + \frac{10u}{24 + 5u}\right) - \left(2 + \frac56 u\right) + \frac{1}{12(2 + \frac56 u)}.$$ It is not difficult to prove that $F(u)$ is strictly increasing on $(0, 1]$.

Fact 4: For all $x\in (0, 1/5]$, $$x\ln x \le \frac{(10 - 5\ln 5)x^2 - (2 + \ln 5)x}{5x + 1}.$$

Using Fact 4, it suffices to prove that $$\frac{12x^3 - x^2 - 14x}{12x^2 + 36x + 24} \ge F\left(\frac{(10 - 5\ln 5)x^2 - (2 + \ln 5)x}{5x + 1}\right)$$ which is true (simply a polynomial inequality).