Does it follow for $x \ge 785$, that Gautschi's Inequality implies that $\frac{\Gamma(2x + 3 - \frac{1.25006}{\ln n})}{\Gamma(2x+1)} > x^2$ Does it follow for $x \ge 785$, that Gautschi's Inequality implies that $\frac{\Gamma(2x + 3 - \frac{1.25006}{\ln n})}{\Gamma(2x+1)} > x^2$
Here's my reasoning.  Please let me know if I made any mistakes or made any jumps in my logic.
(1)  From Gautschi's Inequality, from any real $z$ and any real $s$ where $0 < s < 1$, it follows that:
$$z^s > \frac{\Gamma(z+s)}{\Gamma(z)} > (z)(z+1)^{s-1}$$
(2) Setting $z = 2x+2$ gives us:
$$(2x+2)^s > \frac{\Gamma(2x+2+s)}{\Gamma(2x+2)} > (2x+2)(2x+3)^{s-1}$$
(3) Multiplying $2x+1$ to both sides:
$$(2x+1)(2x+2)^s > \frac{\Gamma(2x+2+s)}{\Gamma(2x+1)} > (4x^2+6x+2)(2x+3)^{s-1}$$
(4) Since $\dfrac{1.25506}{\ln x} < 1$ for $x \ge 4$, setting $s = 1 - \dfrac{1.25506}{\ln x}$ gives us:
$$\frac{\Gamma(2x+3-\frac{1.25506}{\ln x})}{\Gamma(2x+1)} > (4x^2+6x+2)(2x+3)^{-\frac{1.25506}{\ln x}} = \frac{4x^2+6x+2}{(2x+3)^{\frac{1.25506}{\ln x}}}$$
(5)  Since for $x \ge 785$ (see here for details), $(2x+3)^{\frac{1.25506}{\ln x}} < 4$, it follows that for $x \ge 785$:
$$\frac{\Gamma(2x+3-\frac{1.25506}{\ln x})}{\Gamma(2x+1)} > x^2$$

I have not found any mistakes.

In the following, let us prove, with the help of WolframAlpha, that if $x \ge 785$, then $(2x+3)^{\frac{1.25506}{\ln x}} < 4$ since it seems that you forgot to show the details.
The inequality is equivalent to $f(x)\gt 0$ where
$$ f(x)=(\ln 4)(\ln x)-1.25506\ln(2x+3)$$
with
$$f'(x)=\frac{(\ln(16)-2.51012)x+3\ln 4}{x(2x+3)}\gt 0$$
since $\ln(16)-2.51012\gt 0\ $ (see here).
Since $f(x)$ is increasing with
$$f(785)=(\ln 4)(\ln 785)-1.25506\ln(1573)\gt 0$$
(see here), it follows that $f(x)\gt 0$ for $x\ge 785$.
