Using $\sqrt{1-t}\leq 1-\frac t2$ to show that $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\geq\frac1{2\sqrt n}$

Question

I have a problem that tells me to use that $\sqrt{1-t}\leq 1-\frac t2$ for $t\in(0,1)$ to show by induction that $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\geq\frac1{2\sqrt n}$

So far I have shown that $\frac{1\cdot3\cdot5\cdots(2n-1)}{1\cdot2\cdot3\cdots n}\leq 2^n$ from a previous problem. I don't know if that can be used. But the first premise has to be used.

Answer 1

Write the product

$$\prod_{k=1}^n \frac{2k-1}{2k} = \prod_{k=1}^n \left( 1 - \frac{1/k}{2}\right)$$

and use the given inequality for $k > 1$ to get the inequality

$$\prod_{k=1}^n \frac{2k-1}{2k} \geqslant \left(1-\frac{1}{2}\right)\prod_{k=2}^n \sqrt{1-\frac{1}{k}} = \frac{1}{2}\prod_{k=2}^n\sqrt{\frac{k-1}{k}}.$$