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 $x \ge 785$:
$$\frac{\Gamma(2x+3-\frac{1.25506}{\ln x})}{\Gamma(2x+1)} > x^2$$
 A: 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 $x\ge 785$.
A: It seems to me that the inequality holds even for smaller values of $x$.
Considering that we look for the zero of function
$$f(x)=\log \left(\frac{\Gamma \left(2x+3-\frac{a}{\log (x)}\right)}{\Gamma (2
   x+1)}\right)-2\log(x)$$ and using Stirling approximation plus Taylor series
$$f(x)=-\left(\frac{a \log (2)}{\log (x)}+a-2\log (2)\right)+\frac{(3 \log (x)-a) (2 \log
   (x)-a )}{4 x \log ^2(x)}+O\left(\frac{1}{x^2}\right)$$  an overestimate of the solution is given by
$$2^{-\frac{a}{a-2 \log (2)}}$$ which, for $a=1.25506$ gives $756.66$.
Newton iterates are
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 756.000 \\
 1 & 455.485 \\
 2 & 512.524 \\
 3 & 517.233 \\
 4 & 517.260
\end{array}
\right)$$
