Prove $B(a,\frac{1}{2}) \ge \frac{1}{a}$ for $a \ge 1$

Question

I would like to prove

$B(a,\frac{1}{2}) \geq \frac{1}{a}$ for $a \geq 1$,

where B represents the Beta function.

It seems to be true, but I haven't found a nice proof yet. Tried to work with the integral formula but I couldn't get a clean picture.

Does anyone have an idea on how to prove this?

Here is a start with the Gamma function.

\begin{align*} B(a,\frac{1}{2}) = \frac{\Gamma(a) \cdot \Gamma(\frac{1}{2})}{\Gamma(a+\frac{1}{2})} \geq \frac{1}{a} \end{align*}

Now I multiply both sides with $a$.

\begin{align*} a \cdot \frac{\Gamma(a) \cdot \Gamma(\frac{1}{2})}{\Gamma(a+\frac{1}{2})} \geq 1 \end{align*}

Now I use the fact that $\Gamma(a+b) \leq \Gamma(a+c)$ for $b \leq c$.

\begin{align*} a \cdot \frac{\Gamma(a) \cdot \Gamma(\frac{1}{2})}{\Gamma(a+\frac{1}{2})} \geq a \cdot \frac{\Gamma(a) \cdot \Gamma(\frac{1}{2})}{\Gamma(a+1)} = a \cdot \frac{\Gamma(a) \cdot \Gamma(\frac{1}{2})}{a \cdot \Gamma(a)} = \Gamma(\frac{1}{2}) \geq 1 \end{align*}

Answer 1

For $a\geq 1$, we have\begin{align*} B\left(a,\frac{1}{2}\right)&=\int_0^1\frac{t^{a-1}}{\sqrt{1-t}}dt\\ &\geq\int_0^1t^{a-1}dt\\ &=\frac{1}{a}. \end{align*}