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...

I was composing the following when the original answer by @MikeSpivey appeared; his edited answer provides considerably more information. The short calculation below arrives at the desired result without requiring integration by parts but does need knowledge of the relationship between the Beta and Gamma functions.
For $c \geq 2$
$$\begin{align*}
\text{Beta}(c+1, 1/c) &= \frac{\Gamma(c+1)\Gamma(1/c)}{\Gamma(c + 1 + 1/c)}\\
&= \frac{c}{c+1/c}\times \frac{c-1}{c-1+1/c}\times \cdots 
\times \frac{2}{2+1/c}\times \frac{1}{1+1/c} \times \frac{1}{(1/c)\Gamma(1/c)}\times \Gamma(1/c).
\end{align*}$$
The product of the last three terms on the right is $c^2/(c+1)$ while all the other terms have value less than $1$.  Hence,
$$\text{Beta}(c+1, 1/c) < \frac{c^2}{c+1} \Rightarrow \frac{1}{c}\text{Beta}(c+1, 1/c) < \frac{c}{c+1} = 1 - \frac{1}{c+1}$$
