I am trying to find an upper and lower bound for the following function:
$$f(x) = \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$
where
$$\frac{1}{b_1} = \frac{1}{b_2} + \frac{1}{b_3}$$
Is it always the case:
$\ln\Gamma(\frac{x}{b_1}) - \ln\Gamma(\frac{x}{b_2}+\frac{1}{2}) - \ln\Gamma(\frac{x }{b_3}+\frac{1}{2}) \le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!) \le \ln\Gamma(\frac{x}{b_1}+1) - \ln\Gamma(\frac{x}{b_2} + \frac{1}{2}) - \ln\Gamma(\frac{x}{b_3}+\frac{1}{2})$
If you can show analysis behind your answer, that will really help.
My main interest is the analysis behind the answer.
Edit: I removed the attempt at a partial answer which was incorrect. I posted the partial answer as a separate question here.
The answer to my question above is no. It is true when $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} \ge 1$.