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

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 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 ?

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 $$x=0.103099$$.