About the use of Stirling approximation

Question

How to prove this inequality: $$\ln \Gamma \left( x \right)-2\ln \Gamma \left( \frac{x+1}{2} \right)>\frac{2x}{3}$$ Sry I forgot to mention that $x>300$

Answer 1

Another approach would be to use approximations. There is a quickly convergent version of Stirling's formula which goes like this:$$\ln{\Gamma(x)}=\left(x-\frac{1}{2}\right)\ln{x}-x+\frac{\ln{2\pi}}{2}+\frac{1}{12(x+1)}+O(x^{-2})$$ (see http://goo.gl/9hsnO). Derive upper and lower bound from this and plug back into your inequality. You'll end up with a relatively straightforward logarithmic inequality.

Answer 2

Simply take the derivative of your function: $$f'(x) = \psi(x)-\psi\left(\frac{x+1}{2}\right)$$ And then show that $f''(x) >0$ and that $f'(x)>2/3$ for say $x=20$. This shows that the derivative is larger for all $x>20$.

Now calculate $f(300) > 200$ to prove that $f(x) > 2x/3$ for all $x>300$.