Bound on the Beta function For positive integers x and y, we have that
$$
B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} = \frac{1}{x} \left( \begin{array}{c} x+y-1 \\ x \end{array} \right)^{-1} .
$$
However,
$$
\left( \begin{array}{c} a+b \\ a \end{array} \right) = \left(\frac{b+1}{1}\right)\left(\frac{b+2}{2}\right) ... \left(\frac{b+a}{a}\right) \leq (b+1)^a .
$$
Therefore, using $a=x$ and $b=y-1$, we have that
$$
B(x,y) \geq \frac{y^{-x}}{x}.
$$
Is the final inequality true for real values $x \geq 1$ and $y\geq 1$? If so, how is this proved?
 A: OK, I finally figured it out. This is for integer $x\geq 1$ and real valued $y \geq 1$, which is actually all that I needed. Note that you can swap x and y due to the symmetry of the beta function. I believe that the inequality should hold when neither $x$ nor $y$ are integers (but still at least 1), but I did not actually need this result.
We have that
\begin{equation*}
 B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.
\end{equation*}
However, note that $\Gamma(x) = (x-1)! = \frac{x!}{x}$ for integers $x \geq 1$. We also have that
$$
\begin{split}
\Gamma(x+y) & = (x+y-1) \Gamma(x+y-1) \\
&= (x+y-1)(x+y-2) \Gamma(x+y-2) \\
&= \cdots \\
&= \left({\prod_{j=1}^x (x+y-j)}\right) \Gamma(x+y-x) \\
&= \Gamma(y) \left({\prod_{j=1}^x (y-1+j)}\right) .
\end{split}
$$
Putting all of this together, we have that (Equation (1))
$$
\begin{split}
B(x,y) &= \frac{x! \Gamma(y)}{x \Gamma(y) \left({\prod_{j=1}^x (y-1+j)}\right) } \\
&= \frac{1}{x} \frac{x!}{\left({\prod_{j=1}^x (y-1+j)}\right) } \\
&= \frac{1}{x\left({\prod_{j=1}^x \frac{y-1+j}{j} }\right) } .
\end{split}
$$
However, for $j\geq 1$, note that
$$
\begin{split}
y \geq 1 &\implies y-1 \geq 0 \\
&\implies (y-1)(j-1) \geq 0 \\
&\implies 1 - y - j + jy \geq 0 \\
&\implies y-1+j \leq jy \\
&\implies \frac{y-1+j}{j} \leq y .
\end{split}
$$
We also have that $0 < \frac{y-1+j}{j}$, and therefore we have that
$$
{\prod_{j=1}^x \frac{y-1+j}{j} } \leq \prod_{j=1}^x y = y^x .
$$
Combining this inequality with Equation (1), we have that
$$
B(x,y) \geq \frac{y^{-x}}{x} . \phantom{mmmm}\blacksquare
$$
