I am struggling with the inequality below: $$\ln\Gamma(x+1) - 2\ln\Gamma\left(\frac 12 x + \frac 12\right)\geq \ln(\left\lfloor x \right\rfloor !)-2\ln\left(\left\lfloor \frac x2 \right\rfloor !\right)$$ I know that $$\ln\Gamma(x+1)\geq\ln(\left\lfloor x \right\rfloor !)$$ so that part is easy to get. However, I am unsure of how to proceed with the next term. Any help is appreciated! Additionally, what about the flip side?

$$\ln\Gamma(x) - 2\ln\Gamma\left(\frac 12 x + \frac 12\right)\leq \ln(\left\lfloor x \right\rfloor !)-2\ln\left(\left\lfloor \frac x2 \right\rfloor !\right)$$ How would I prove that?


After exponentiation we get:

$$ \frac{\Gamma(x+1)}{\Gamma(\frac{x}{2}+\frac{1}{2})^2}\ge \frac{\Gamma(\lfloor x\rfloor +1)}{\Gamma(\lfloor \frac{x}{2}\rfloor +1)^2} $$

it gets directly obvious, since : $$ {\Gamma(x+1)}\ge {\Gamma(\lfloor x\rfloor +1)}$$

for $x>0$,

and also:

$$ \Gamma(\frac{x}{2}+\frac12) \le \Gamma(\lfloor \frac{x}{2}\rfloor +1\rfloor).$$

for $x>0$.

  • $\begingroup$ I added an edit above. The same question had a part 2, which I am also confused about. Can you please take a look and see what you can do? $\endgroup$ – D.R. Mar 27 '18 at 23:36

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