Minimize the coefficient of asymptotic normality I need to minimize this complicated function: 
$$\dfrac{\Gamma(2x+1)-\Gamma(x+1)^2}{x^2 \Gamma(x+1)^2}$$ (For people wondering where this function came from — it's the (multiple of) coefficient of asymptotic normality of the estimator $\widehat{\alpha}$ for parameter $\alpha$ of exponential distribution obtained by method of moments using the sample function $t^x$ ($x$ is parameter here)).
Since the coefficient of asymptotic normality must be positive, I need to find the minimum only for $x > -\dfrac{1}{2}$. Simple analysis shows that minimum exists, because function tends to infinity both at $x \rightarrow -\dfrac{1}{2}$ and $x \rightarrow \infty$.
Consideration of only discrete arguments indicates that $\dfrac{(2k)! - (k!)^2}{k^2 (k!)^2}$ attains its minimum at $k = 1$. The graph also shows that $1$ has to be a minimum of this function.
Is there a way to prove it analytically? Thank you in advance.
 A: I can show that the derivative f the above expression at $x=1$ is zero.
Rewrite that expression about $y=x-1$ and get
$$f(y) = \frac{1}{(y+1)^2} \left ( \frac{\Gamma(2 y+3)}{\Gamma(y+2)^2}-1\right)$$
$$f'(y) = -\frac{2}{(y+1)^3} \left ( \frac{\Gamma(2 y+3)}{\Gamma(y+2)^2}-1\right) + \frac{2}{(y+1)^2} \left ( \frac{\Gamma(y+2)^2 \Gamma'(2 y+3)-\Gamma(y+2)\Gamma'(y+2)\Gamma(2 y+3)}{\Gamma(y+2)^4}\right ) $$
$$f'(0) = -2+2(\Gamma'(3)-2 \Gamma'(2))$$
Now, for any positive integer $n$,
$$\Gamma'(n) = \Gamma(n) (H_{n-1}-\gamma)$$
where $\gamma$ is Euler's gamma and $H_n$ is the $n$th term of the Harmonic series.  Therefore
$$\Gamma'(2) = \Gamma(2) (1-\gamma) = 1-\gamma$$
$$\Gamma'(3) = \Gamma(3) (1+(1/2)-\gamma) = 3-2 \gamma$$
Putting it all together:
$$f'(0) = -2 + 2 (3-2 \gamma -2 + 2 \gamma) = 0$$
Now, to prove that $y=0$ is a minimum, we need to evaluate $f''(0)$, which I found to be too painful to do by hand for now.  Mathematica, however, returns a value of 
$$f''(0) = \frac{2 \pi^2}{3}-6 >0$$
so yes, this is a minimum as observed.
A: Don't really know would this be of any help, but you can put it like this:
You can use the following properties: 
$\Gamma(x+1)=x\Gamma(x)$ and $\Gamma(2x+1)=2x\Gamma(2x)=2x\frac{2^{2x-1}}{\sqrt{\pi}}\Gamma(x)\Gamma(x+\frac{1}{2})$
So after cancelling out $x^2\Gamma(x)$ terms this will lead to:
$\frac{2^{2 x} \Gamma \left(x+\frac{1}{2}\right)}{\sqrt{\pi } x^3 \Gamma (x)}-\frac{1}{x^2}$
At this point I don't see how it solves the problem, but at least it reduces the amount of gamma-functions. :)
The graph shows that the minimum will be attained at the point $x=1$.
 And numeric minimization prooves it.
