This is the second attempt at a proof. My first attempt had a flaw in its logic.
After reviewing the mistake in logic, I believe that with a revised logic, the argument can be saved.
The revision consists of two arguments. The argument presented below covers the condition where $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} < 1$. My question here covers the condition where $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} \ge 1$
I am very glad to be corrected if someone is able to find a mistake. :-)
This is an attempt to generalize one of the steps in Ramanujan's proof of Bertrand's postulate.
In particular, Ramanujan's does the following comparison in step (8):
$$\ln\Gamma(x) - 2\ln\Gamma(\frac{x}{2} + \frac{1}{2}) \le \ln(\lfloor{x}\rfloor!) - 2\ln(\lfloor\frac{x}{2}\rfloor!)$$
It occurs to me that this can be generalized to:
$$\ln\Gamma(\frac{x}{b_1}-\frac{3}{16}) - \ln\Gamma(\frac{x}{b_2}+\frac{19}{32}) - \ln\Gamma(\frac{x}{b_3}+\frac{19}{32})\le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$
when:
$$\frac{x}{b_1} = \frac{x}{b_2} + \frac{x}{b_3}$$
and:
$$\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} < 1$$
where:
$$x > 36$$
Here's the argument for this generalization:
Let:
$$\{\frac{x}{b_i}\} = \frac{x}{b_i} - \lfloor\frac{x}{b_i}\rfloor$$
where:
$$0 \le \{\frac{x}{b_i}\} < 1$$
Since:
$$\{\frac{x}{b_1}\} + \lfloor\frac{x}{b_1}\rfloor = \{\frac{x}{b_2}\} + \lfloor\frac{x}{b_2}\rfloor + \{\frac{x}{b_3}\} + \lfloor\frac{x}{b_3}\rfloor$$
We have:
$$\{\frac{x}{b_1}\} = \{\frac{x}{b_2}\} + \{\frac{x}{b_3}\}$$
and
$$\lfloor\frac{x}{b_1}\rfloor = \lfloor\frac{x}{b_2}\rfloor + \lfloor\frac{x}{b_3}\rfloor$$
If $\Delta{t_1} \ge \Delta{t_2} + \Delta{t_3}$ and $x_1 + \Delta{t_1} \ge x_2 + \Delta{t_2} \ge x_3 + \Delta{t_3} > 0$,
Using the logic in the answer here:
$$\frac{\Gamma(x_1 + \Delta{t_1})}{\Gamma(x_1)} \ge \frac{\Gamma(x_2 + \Delta{t_2})}{\Gamma(x_2)}\frac{\Gamma(x_3 + \Delta{t_3})}{\Gamma(x_3)}$$
Let:
$x_1 = \frac{x}{b_1}-\frac{3}{16}$, $\Delta{t_1} = \frac{19}{16}-\{\frac{x}{b_1}\}$,
$x_2 = \frac{x}{b_2}+\frac{19}{32}$, $\Delta{t_2} = \frac{13}{32}-\{\frac{x}{b_2}\}$
$x_3 = \frac{x}{b_3}+\frac{19}{32}$, $\Delta{t_3} = \frac{13}{32}-\{\frac{x}{b_3}\}$
where $\frac{x}{b_2} \ge \frac{x}{b_3}$ (Otherwise, switch the two values).
Then:
$$\frac{\Gamma(\lfloor\frac{x}{b_1}\rfloor+1)}{\Gamma(\frac{x}{b_1}-\frac{3}{16})} \ge \frac{\Gamma(\lfloor\frac{x}{b_2}\rfloor+1)}{\Gamma(\frac{x}{b_2}+\frac{19}{32})}\frac{\Gamma(\lfloor\frac{x}{b_3}\rfloor+1)}{\Gamma(\frac{x}{b_3}+\frac{19}{32})}$$
So then it follows:
$$\ln\Gamma(\lfloor\frac{x}{b_1}\rfloor + 1) - \ln\Gamma(\frac{x}{b_1} - \frac{3}{16}) \ge \ln\Gamma(\lfloor\frac{x}{b_2}\rfloor+1) - \ln\Gamma(\frac{x}{b_2}+\frac{19}{32}) + \ln\Gamma(\lfloor\frac{x}{b_3}\rfloor + 1) - \ln\Gamma(\frac{x}{b_3}+\frac{19}{32})$$
And we have shown:
$$\ln\Gamma(\frac{x}{b_1}-\frac{3}{16}) - \ln\Gamma(\frac{x}{b_2}+\frac{19}{32}) - \ln\Gamma(\frac{x}{b_3}+\frac{19}{32}) \le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$
Here's an example:
Let $x=77.1$, $b_1=6$, $b_2=7$, $b_3=42$
So that:
$\{\frac{77.1}{7}\} \approx 0.0143$, $\{\frac{77.1}{42}\} \approx 0.8357$
Then:
$$\ln\Gamma(\frac{77.1}{6}-\frac{3}{16}) - \ln\Gamma(\frac{77.1}{7}+\frac{19}{32}) - \ln\Gamma(\frac{77.1}{42} + \frac{19}{32}) \le \ln(\lfloor\frac{77.1}{6}\rfloor!) - \ln(\lfloor\frac{77.1}{7}\rfloor!) - \ln(\lfloor\frac{77.1}{42}\rfloor!)$$
Please let me know if you see any mistakes.
Thanks,
-Larry
Note 1: I came up with $\frac{x}{b_1}-\frac{3}{16}$ by comparing $\frac{x}{b_1}-1$ and then comparing $\frac{x}{b_1}-\frac{1}{2}$. I found that by taking consistent averages and satisfying these conditions:
If $\Delta{t_1} \ge \Delta{t_2} + \Delta{t_3}$ and $x_1 + \Delta{t_1} \ge x_2 + \Delta{t_2} \ge x_3 + \Delta{t_3} > 0$,
I was able to get a tighter lower bound. I should be able to improve upon $\frac{x}{b_1} - \frac{3}{16}$ if I wanted to.
Note 2: I have not been able to show that this tightening of the lower bound works in all cases. This is a gap in my argument. I still need to show that using $\frac{x}{b_1}-\frac{3}{16}$ is always an improvement to using the more obvious $\frac{x}{b_3}-1$. My argument shows that it will work but not that it is better.
Looking at a graph it seems plausible that your result holds for $x>36$, but it does not hold for $35\le x <36$ where the right side is zero.
The flaw in this case is that $\{x/b_3\}>13/16$, allowing $\Delta t_3<0$ and $\Delta t_2>\Delta t_1$, violating the conditions for the $x_i+\Delta t_i$ inequality.