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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 $x>300$

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Try plugging in $x=1$ to see what happens. –  Byron Schmuland Jan 28 '13 at 13:41
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@gauss115 - please revise the question, as Byron pointed out the statement as it stands is false. –  nbubis Jan 28 '13 at 13:54
    
@nbubis sry! The condition is $x>300$ –  gauss115 Jan 28 '13 at 14:00
    
How do you know this is true? What did you try to show this is true? –  Did Jan 28 '13 at 14:06

2 Answers 2

up vote 1 down vote accepted

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.

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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 $x>300$.

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