I am trying to show that the following function has a lower bound of $\ \frac{1}{2}$ for all $c\geq 2$. Or, alternatively, that that function increases with $\ c$:
$\frac{\Gamma\left[1+\frac{1}{c}\right] \Gamma[1+c]}{\Gamma\left[1+\frac{1}{c}+c\right]}$
Note that this can be rewritten a few ways, such as $\ c\text{ B}[1+\frac{1}{c}, c]$ Plotting it out to 10,000 or so, it seems very clear that it is increasing, approaching 1, but I can't figure out how to prove it either through integration or properties of the special functions. Can anyone help me out? Thank you!
Note: I found in this document in Theorem 3, if $(a-1)(b-1) \geq 0$ then $\text B(a,b) \geq \frac{1}{ab}$
So, I thought that if I took the function as $\ c \text B(1+ \frac{1}{c},c)$ where $\ (\frac{1}{c})(c-1) \geq 0$ we have $c \text B(1+ \frac{1}{c},c) \geq \frac{c}{\frac{c+1}{c} c} = \frac{c}{c+1} > \frac{1}{2}$
However, I must be doing something wrong, as the value of $\ c \text B(1+ \frac{1}{c},c)$ when $c=2$ is actually less than $\frac{2}{3}$ (though greater than $\frac{1}{2}$)