About the inequality $\left(\ln\left(\frac{4+x}{2x+2}\right)\right)^{2}\leq x!$

Problem :

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

$$\left(\ln\left(\frac{4+x}{2x+2}\right)\right)^{2}\leq x!\tag{I}$$

My attempt:

I have tried :

$$x!=\frac{1}{x+1}-\gamma+O(x^2)$$

Wich is not enought .

Also I have tried derivative without success unfortunately .I think we need more terms in the power series like :

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

The problem is we get an upper bound not an lower bound ...

Motivation

My motivation is to find a sufficiently accurate inverse of the Gamma function : see here

Question :

How to show $(I)$ ?

Thanks.

With the substitution $x = y - 1$, taking logarithm, it suffices to prove that, for all $0 < y < 1$, $$\ln \Gamma(y) \ge 2\ln \ln \frac{3 + y}{2y}.$$

Using the known property $$-\ln \Gamma(y) = \ln y + \gamma y + \sum_{n=1}^\infty [\ln(1 + y/n) - y/n],$$ we have, for all $0 < y < 1$, $$\ln \Gamma(y) \ge - \ln y - \gamma y + \sum_{n=1}^2 [y/n - \ln(1 + y/n)]$$ where we have used $u \ge \ln (1 + u)$ for all $u\ge 0$.

It suffices to prove that, for all $0 < y < 1$, $$- \ln y + (3/2 - \gamma)y - \ln(1 + y) - \ln(1 + y/2) \ge 2\ln \ln \frac{3 + y}{2y}$$ or $$(3/2 - \gamma)y - \ln(1 + y) - \ln(1 + y/2) \ge 2\ln \left(\sqrt{y}\ln \frac{3 + y}{2y}\right).$$

Fact 1: For all $0 < y < 1$, $$\sqrt{y}\ln \frac{3 + y}{2y} \le \frac{7770\gamma + 47353}{195804\gamma - 59850} - \frac{8}{63}y.$$ (Hint: Use $y = z^2$ and take derivative.)

Using Fact 1, it suffices to prove that $$(3/2 - \gamma)y - \ln(1 + y) - \ln(1 + y/2) \ge 2\ln \left(\frac{7770\gamma + 47353}{195804\gamma - 59850} - \frac{8}{63}y\right).$$ Denote $f(y) = \mathrm{LHS} - \mathrm{RHS}$. We have $$f'(y) = (16y - 5)g(y).$$ where $$g(y) = \frac{(3108\gamma^2 - 5612\gamma + 1425)y^2 + (9324\gamma^2 - 10620\gamma + 9799)y + 6216\gamma^2 - 1900\gamma + 12160}{[(-49728\gamma + 15200)y + 15540 \gamma + 94706](2 + y)(1 + y)}.$$ It is easy to prove that $g(y) > 0$ for all $0 < y < 1$. Thus, we have $f'(5/16) = 0$, and $f'(y) < 0$ for all $y\in (0, 5/16)$, and $f'(y) > 0$ for all $y\in (5/16, 1)$. Thus, for all $0 < y < 1$, $$f(y) \ge f(5/16) > 0.$$

We are done.

In this range, showing that $$\Bigg[\log \left(\frac{x+4}{2 x+2}\right)\Bigg]^2 \le x!$$ seems to be the same as showing that $$\log \left(\frac{x+4}{2 x+2}\right)\le \sqrt{x!}$$ Consider the function $$f(x)=\sqrt{x!}-\log \left(\frac{x+4}{2 x+2}\right)$$ $$f'(x)=\frac{3}{x^2+5 x+4}+\frac{1}{2} \sqrt{\Gamma (x+1)} \,\psi ^{(0)}(x+1)$$ $$f''(x)=-\frac{1}{(x+1)^2}+\frac{1}{(x+4)^2}+\frac{1}{4} \sqrt{\Gamma (x+1)} \left(\psi ^{(0)}(x+1)^2+2 \psi ^{(1)}(x+1)\right) $$ By inspection $f'(x)=0$ close to $x=-\frac 7{10}$. For this value $$f\left(-\frac{7}{10}\right)=\sqrt{\Gamma \left(\frac{3}{10}\right)}-\log \left(\frac{11}{2}\right) >0$$ $$f''\left(-\frac{7}{10}\right)=\frac{1}{4} \sqrt{\Gamma \left(\frac{3}{10}\right)} \left(\psi ^{(0)}\left(\frac{3}{10}\right)^2+2 \psi ^{(1)}\left(\frac{3}{10}\right)\right)-\frac{4000}{363} > 0$$ So, we have the minimum $$f\left(-\frac{7}{10}\right)=\sqrt{\Gamma \left(\frac{3}{10}\right)}-\log \left(\frac{11}{2}\right)=0.0248672$$ A full minimization of $f(x)$ gives a value of $0.02486720$ at $x=-0.700265$.