Does it follow that for $x > e^3$, $\dfrac{\Gamma\left(2x+2 - \frac{1.25506(x+1)}{\ln(x+1)}\right)}{\Gamma\left(2x - \frac{1.25506(x)}{\ln x}\right)} > \dfrac{\Gamma\left(x+8 - \frac{1.25506}{\ln x}\right)}{\Gamma(x+6)}$
Note: My interest in $\dfrac{1.25506x}{\ln x}$ comes from this upper bound for the prime counting function:
$$\pi(x) < \dfrac{1.25506x}{\ln x}$$
Here is my thinking:
(1) $f(x) = \dfrac{1.25506x}{\ln x}$ is an increasing function at $x > e$ since:
$f'(x) = \dfrac{1.25506\ln(x) - 1.25506}{(\ln x)^2}$ is positive when $\ln(x) > 1$
It follows that: $\dfrac{1.25506(x+1)}{\ln (x+1)} - \dfrac{1.25506x}{\ln x}> 0$
(2) $\dfrac{1.25506(x+1)}{\ln (x+1)} - \dfrac{1.25506x}{\ln x} < \dfrac{1.25506}{\ln x}$ since:
$$\dfrac{1.25506(x+1)}{\ln (x+1)} - \dfrac{1.25506x}{\ln x} < \dfrac{1.25506(x+1)}{\ln(x)} - \dfrac{1.25506x}{\ln x} = \dfrac{1.25506}{\ln x}$$
(3) From the log convex property of the Gamma function, does it follows that for all positive real $a,b,c$
$$\dfrac{\Gamma(a+b))}{\Gamma(a)} < \dfrac{\Gamma(a+b+c)}{\Gamma(a+c)}$$
(4) if $x > e^3$, then $2x - \dfrac{1.25506(x+1)}{\ln(x+1)} > x+6$
- $\ln(x+1) > 1.25506 + 1$
- $x\ln(x+1) > 1.25506x + x > 1.25506(x+1) + (x-1.25506)$
- $x > \dfrac{1.25506(x+1)}{\ln(x+1)} + \dfrac{x-1.25506}{\ln(x+1)}> \dfrac{1.25506(x+1)}{\ln(x+1)} + 6$
- $2x - \dfrac{1.25506(x+1)}{\ln(x+1)} > x+6$
(5) From step(2) and (3), for $x \ge e^2$, it follows that if $s = 2-\left(2x+2 - \frac{1.25506(x+1)}{\ln(x+1)}\right) + \left(2x - \frac{1.25506(x)}{\ln x}\right)$, then:
$1 < s < 2$
$\dfrac{\Gamma\left(2x+2 - \frac{1.25506(x+1)}{\ln(x+1)}\right)}{\Gamma\left(2x - \frac{1.25506(x)}{\ln x}\right)} = \dfrac{\Gamma\left(2x - \frac{1.25506(x)}{\ln(x)}+s\right)}{\Gamma\left(2x - \frac{1.25506(x)}{\ln x}\right)} > \dfrac{\Gamma\left(2x - \frac{1.25506(x+1)}{\ln(x+1)}+s\right)}{\Gamma\left(2x - \frac{1.25506(x+1)}{\ln (x+1)}\right)} > \dfrac{\Gamma\left(2x - \frac{1.25506(x+1)}{\ln(x+1)}+2 - \frac{1.25506}{\ln x}\right)}{\Gamma\left(2x - \frac{1.25506(x+1)}{\ln (x+1)}\right)}$
(6) From step(4) and step(3):
$\dfrac{\Gamma\left(2x - \frac{1.25506(x+1)}{\ln(x+1)}+2 - \frac{1.25506}{\ln x}\right)}{\Gamma\left(2x - \frac{1.25506(x+1)}{\ln (x+1)}\right)} > \dfrac{\Gamma\left(x+8 - \frac{1.25506}{\ln x}\right)}{\Gamma(x+6)}$
It seems to be true even for $x=e$.
Let $x=e+\epsilon$ and developing as Taylor series around $\epsilon=0$, the numerical evalution of the terms gives $$\log \left(\frac{\text{lsh}}{\text{rhs}}\right)=0.0337736+0.189071 \epsilon +O\left(\epsilon ^2\right)$$ which is zero if $\epsilon=-0.178629$ that is to say $x=2.53965$.
Using again Newton method, the iterates are $$\left( \begin{array}{cc} n & x_n \\ 0 & 2.53965 \\ 1 & 2.57365 \\ 2 & 2.57590 \\ 3 & 2.57591 \end{array} \right)$$