Is it possible to generalize Ramanujan's lower bound for factorials that he used in his proof of Bertrand's Postulate? I am starting to feel more confident in my understanding of Ramanujan's proof of Bertrand's postulate.  I hope that I am not getting overconfident.
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:
$$\ln\Gamma(\frac{x}{b_1}) - \ln\Gamma(\frac{x}{b_2} + \frac{1}{2}) - \ln\Gamma(\frac{x}{b_3} + \frac{1}{2})\le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$
when:
$$\frac{x}{b_1} = \frac{x}{b_2} + \frac{x}{b_3}$$
I would really appreciate it if my argument could be reviewed and someone could call out any mistakes either in the answer or in comments.  :-)
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: 
$$\{\frac{x}{b_1}\} \le \{\frac{x}{b_2}\} + \{\frac{x}{b_3}\}$$
So that:
$$-\{\frac{x}{b_1}\} \ge -\{\frac{x}{b_2}\} + -\{\frac{x}{b_3}\}$$
$$2-\{\frac{x}{b_1}\} \ge 1-\{\frac{x}{b_2}\} + 1-\{\frac{x}{b_3}\}$$
$$1-\{\frac{x}{b_1}\} \ge 1-\{\frac{x}{b_2}\} - \frac{1}{2} + 1-\{\frac{x}{b_3} \} - \frac{1}{2}$$
$$\lfloor\frac{x}{b_1}\rfloor + 1 - \frac{x}{b_1} \ge (\lfloor\frac{x}{b_2}\rfloor + 1 - \frac{x}{b_2} - \frac{1}{2}) + (\lfloor\frac{x}{b_3}\rfloor + 1 - \frac{x}{b_3} - \frac{1}{2})$$
$$\lfloor\frac{x}{b_1}\rfloor + 1 \ge (\lfloor\frac{x}{b_2}\rfloor + 1) + (\lfloor\frac{x}{b_3}\rfloor + 1)$$
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:
$x_1 = \frac{x}{b_1}$,  $\Delta{t_1} = 1 - \{\frac{x}{b_1}\}$,
$x_2 = \frac{x}{b_2}+\frac{1}{2}$,  $\Delta{t_2} = \frac{1}{2} - \{\frac{x}{b_2}\}$
$x_3 = \frac{x}{b_3}+\frac{1}{2}$,  $\Delta{t_3} = \frac{1}{2} - \{\frac{x}{b_3}\}$
where $\frac{x}{b_2} \ge \frac{x}{b_3}$ (Otherwise, switch the two values).
Then:
$$\frac{\Gamma(\lfloor\frac{x}{b_1}\rfloor+1)}{\Gamma(\frac{x}{b_1})} \ge \frac{\Gamma(\lfloor\frac{x}{b_2}\rfloor+1)}{\Gamma(\frac{x}{b_2} + \frac{1}{2})}\frac{\Gamma(\lfloor\frac{x}{b_3}\rfloor+1)}{\Gamma(\frac{x}{b_3}+\frac{1}{2})}$$
So then it follows:
$$\ln\Gamma(\lfloor\frac{x}{b_1}\rfloor + 1) - \ln\Gamma(\frac{x}{b_1}) \ge \ln\Gamma(\lfloor\frac{x}{b_2}\rfloor + 1) - \ln\Gamma(\frac{x}{b_2} + \frac{1}{2}) + \ln\Gamma(\lfloor\frac{x}{b_3}\rfloor + 1) - \ln\Gamma(\frac{x}{b_3} + \frac{1}{2})$$
And we have shown:
$$\ln\Gamma(\frac{x}{b_1}) - \ln\Gamma(\frac{x}{b_2}+\frac{1}{2}) - \ln\Gamma(\frac{x }{b_3}+\frac{1}{2}) \le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$
Please let me know if you see any mistakes.
Thanks,
-Larry

Edit: Based on reviewing Zander's answer, I believe that this argument can be saved.  The revision requires two separate arguments:


*

*one for: $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} \ge 1$

*another for: $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} < 1$  


The link for the first argument is here.  The link for the second argument is here.

After this step:
$$
\lfloor\frac{x}{b_1}\rfloor + 1 - \frac{x}{b_1} \ge (\lfloor\frac{x}{b_2}\rfloor + 1 - \frac{x}{b_2} - \frac{1}{2}) + (\lfloor\frac{x}{b_3}\rfloor + 1 - \frac{x}{b_3} - \frac{1}{2})
$$
we have
$$
\lfloor\frac{x}{b_1}\rfloor + 1 \ge (\lfloor\frac{x}{b_2}\rfloor + 1) + (\lfloor\frac{x}{b_3}\rfloor + 1)-1
$$
but what you have is incorrect.
For example, let $x=77.1,b_1=6,b_2=7,b_3=42$, then $\lfloor x/b_1\rfloor=12, \lfloor x/b_2\rfloor=11, \lfloor x/b_3 \rfloor=1$ and $13\not\ge 14$.
And in this case your inequality is also violated:
$$
\ln \Gamma\left(\frac{77.1}{6}\right)-\ln\Gamma\left(\frac{77.1}{7}+\frac{1}{2}\right)-\ln \Gamma\left(\frac{77.1}{42}+\frac{1}{2}\right) = 3.10698\cdots
\\
\ge \ln(12!)-\ln(11!)-\ln(1!) = 2.4849\cdots
$$
