<s>[INST] Beta Function -- finding a lower bound based on parameters I would like to show that 
$$ 1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) \geq \frac{1}{c+1}.$$
for all $c \geq 2$.
I have plotted it out for $c$ up through 200, and it seems to hold.
Does anyone have any tips on if this can be formally shown?  I imagine an expert on the beta function might know...
 [/INST] The definition of the beta function is
$$Beta\left(c+1,\frac{1}{c}\right)  = \int_0^1 t^c (1-t)^{1/c-1} dt.$$ 
Since $c \geq 2$ and $t \in [0,1]$, $t^c \leq t$.  Thus (applying integration by parts to the first integral), we have
$$Beta\left(c+1,\frac{1}{c}\right) \leq \int_0^1 t (1-t)^{1/c-1} dt = \left. -c t (1-t)^{1/c}\right|_0^1 + c\int_0^1 (1-t)^{1/c} dt$$  $$= \left. \frac{-c (1-t)^{1/c+1}}{1+1/c}  \right|_0^1 = \frac{c}{1+1/c} = \frac{c^2}{c+1}.$$
Therefore,
$$1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) \geq 1- \frac{c}{c+1} = \frac{1}{c+1}.$$ 

The bound isn't very tight as $c$ increases, though, as we can see from the asymptotics:
$$1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) = \frac{\log c + \gamma}{c} + O\left(\frac{(\log c)^2}{c^2}\right).$$ 
Derivation:
$$Beta\left(c+1,\frac{1}{c}\right) = \frac{\Gamma(c+1) \Gamma \left(\frac{1}{c}\right)}{\Gamma\left(c+1+\frac{1}{c}\right)} = \frac{c \Gamma(c+1) \Gamma \left(1+\frac{1}{c}\right)}{\Gamma\left(c+1+\frac{1}{c}\right)}.$$
Now, $\Gamma \left(1+\frac{1}{c}\right) = 1 - \frac{\gamma}{c} + O\left(\frac{1}{c^2}\right)$. (This is via the Maclaurin series for $\Gamma(x+1)$; see Expression 8.321 in Gradshteyn and Ryzhik.)
Also (see the DLMF), $$\frac{\Gamma(c+1)}{\Gamma\left(c+1+\frac{1}{c}\right)} = c^{-1/c} \left(1 + O\left(\frac{1}{c^2}\right)\right) = \exp\left(\frac{-\log c}{c}\right)\left(1 + O\left(\frac{1}{c^2}\right)\right)$$
$$= \left(1 - \frac{\log c}{c} + O\left(\frac{(\log c)^2}{c^2}\right)\right)\left(1 + O\left(\frac{1}{c^2}\right)\right) = \left(1 - \frac{\log c}{c} + O\left(\frac{(\log c)^2}{c^2}\right)\right).$$
Putting this all together, we have 
$$1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) = 1 - \frac{c}{c}\left(1 - \frac{\gamma}{c} + O\left(\frac{1}{c^2}\right)\right)\left(1 - \frac{\log c}{c} + O\left(\frac{(\log c)^2}{c^2}\right)\right)$$
$$= \frac{\log c + \gamma}{c} + O\left(\frac{(\log c)^2}{c^2}\right).$$ 
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