Trying to understand a step in Ramanujan's Proof of Bertrand's Postulate regarding the gamma function

Question

My question relates to this step in the proof here:

But it is easy to see that

$$\log \Gamma(x)-2\log\Gamma(\frac12x+\frac12) \le \log\left\lfloor x\right\rfloor!-2\log\left\lfloor\frac12x\right\rfloor!$$

if I use $\left\{\dfrac{x}{2}\right\} = \dfrac{x}{2} - \left\lfloor\dfrac{x}{2}\right\rfloor$ to represent the fractional part.

It is clear to me that the statement is true for when $\left\{\dfrac{x}{2}\right\} \le \dfrac{1}{2}$. I am having trouble understanding why it is necessarily true when $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$

When $\left\{\dfrac{x}{2}\right\} \le \dfrac{1}{2}$, I can use the answer from here.

where

$x_1 = x$, $\Delta{t_1} = 1-\left\{x\right\}$, $x_1+\Delta t_1 = \lfloor{x}\rfloor+1$

$x_2 = \dfrac{x}{2}+\dfrac{1}{2}$, $\Delta{t_2} = \dfrac{1}{2}-\left\{\dfrac{x}{2}\right\}$, $x_2+\Delta t_2 = \lfloor\dfrac{x}{2}\rfloor+1$

$x_3 = \dfrac{x}{2}+\dfrac{1}{2}$, $\Delta{t_3} = \dfrac{1}{2}-\left\{\dfrac{x}{2}\right\}$, $x_3+\Delta t_3 = \lfloor\dfrac{x}{2}\rfloor+1$

to get:

$$\dfrac{\Gamma(\left\lfloor{x}\right\rfloor+1)}{\Gamma({x})} \ge \frac{\Gamma(\left\lfloor\dfrac{x}{2}\right\rfloor+1)}{\Gamma(\dfrac{x}{2}+\dfrac{1}{2})}\dfrac{\Gamma(\left\lfloor\dfrac{x}{2}\right\rfloor+1)}{\Gamma(\dfrac{x}{2}+\dfrac{1}{2})}$$

This approach fails for $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$ since $\Delta{t_2}, \Delta{t_3} < 0$

Can someone provide me the argument for why this inequality is true for the condition where $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$?

Thanks very much.

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Edit: I figured out an argument that works for $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$

I posted it as the answer below.

Answer 1

I figured out an argument that works for $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$

$$\log\Gamma(x) \le \log\Gamma(x + 1 - \left\{ x\right\}) = \log\Gamma(\left\lfloor x\right\rfloor + 1) = \log\left\lfloor x\right\rfloor!$$

$$2\log\Gamma(\dfrac{x}{2} + \dfrac{1}{2}) \ge 2\log\Gamma(\dfrac{x}{2}+1 - \left\{\dfrac{x}{2}\right\}) = 2\log\Gamma(\left\lfloor\dfrac{x}{2}\right\rfloor+1) = 2\log\left\lfloor\dfrac{x}{2}\right\rfloor! $$