A lower bound for the Gamma function : $\Gamma(x)\geq (f(1-x))^x$ on $1\leq x \leq 2$

Question

Claim : Let $1\leq x\leq 2$ then prove or disprove we have :

$$\Gamma(x)\geq (f(1-x))^x$$

Where :

$$f(x)=\left(1+\frac{3}{4}xe^x\right)^{\frac{3}{4}(x+1)}$$

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As attempt we can use Taylor's series of $\Gamma(x)$ around $2$ :

$$\Gamma(x)=1+(1-\gamma)(x-2)+\frac{1}{12}(-12\gamma+6\gamma^2+\pi^2)(x-2)^2+O((x -2)^3)$$

As other option we can take the logarithm and then use derivative but it's quite complicated .In fact it remains to find a bound for the digamma function also see this article about Digamma and gamma function inequalities

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My motivation :

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

Edit :

Second claim :

Following the advice (see ClaudeLeibovici comment)

It seems that the inequality is true as :

$$f(x)=\left(1+\sqrt{\gamma}xe^x\right)^{\sqrt{\gamma}(x+1)}$$

Where $\gamma$ is the Euler-Mascheroni constant.

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Question :

How to (dis)prove the second claim (the first is wrong)?

Answer 1

Let $$g(x)=\big[f(1-x)\big]^x$$

At the end points $$g'(1)=-\frac 9{16} > - \gamma \qquad \text{and} \qquad g'(2)=-\frac{3}{2} \log \left(1-\frac{3}{4 e}\right) > 1- \gamma$$

The derivative of $\Gamma(x)$ cancels very close to $\frac 32$ (more exactly at $x=1.46163$

$$g\left(\frac{3}{2}\right)=\left(1-\frac{3}{8 \sqrt{e}}\right)^{9/16} < \frac{\sqrt{\pi }}{2}=\Gamma\left(\frac{3}{2}\right)$$

If we make a series expansion of $\Gamma(x)-g(x)$ to second order around $x=\frac 32$, the minimum is $x=1.52970$ and, for this value, the difference is $0.0214623$.

A full optimization give a maximum difference of $0.0214623$ at $x=1.52981$.

I refuse to type the analytical expressions.

Edit

As @River Li commented, this inequality does not hold for the range $1 \leq x \leq 1.0336$. Check the derivatives at $x=1$.

Expanding as a series around $x=1$ gives

$$\Gamma(x)-g(x)=\left(\frac{9}{16}-\gamma \right) (x-1)+\frac{\left(-783+768 \gamma ^2+128 \pi ^2\right) }{1536}(x-1)^2+O\left((x-1)^3\right)$$ showing a root at $$x=1+\frac{96 (16 \gamma -9)}{-783+768 \gamma ^2+128 \pi ^2}=1.03070$$

Answer 2

I prefer to add another answer since I am changing the problem.

Let $$f(x)=\left(1+\sqrt{\gamma }\, x\,e^x \right)^{\sqrt{\gamma }\, (x+1)}$$ Notice that $\sqrt{\gamma }=0.759747$ is just $1.3$% larger than the $\frac 34$ in the original question.

As before, let $$g(x)=\big[f(1-x)\big]^x$$ Now, we have $$g'(1)=-\gamma \qquad \text{and} \qquad g'(2)=-2 \sqrt{\gamma } \log \left(1-\frac{\sqrt{\gamma }}{e}\right) > 1-\gamma$$

Around $x=1$ $$\Gamma(x)-g(x)=\frac{1}{12} \left(-12 \gamma +6 \gamma ^{3/2}+\pi ^2\right) (x-1)^2+O\left((x-1)^3\right)$$ and no more root. So, we are fine. $$g\left(\frac{3}{2}\right)=\left(1-\sqrt{\frac{\gamma }{4e}}\right)^{\frac{3 }{4}\sqrt{\gamma }}< \frac{\sqrt{\pi }}{2}=\Gamma\left(\frac{3}{2}\right)$$

If we make a series expansion of $\Gamma(x)-g(x)$ to second order around $x=\frac 32$, the minimum is $x=1.52609$ and, for this value, the difference is $0.0249575$.

A full optimization give a maximum difference of $0.0249575$ at $x=1.52617$.

I still refuse to type the analytical expressions.

Now, the claim is true.