Induction Proof: $\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)} \geq \frac{1}{2n}$

Question

Need help proving with induction that $\displaystyle \frac{1\cdot3\cdot5\cdot7...(2n-1)}{2\cdot4\cdot6\cdot8...\cdot2n} \ge \frac 1{2n}$ for all natural numbers $n$. I just can't even get started with this. Any help appreciated.

I got here: $\frac{1*3*5*...(2n+1)}{2*4*6*...(2n+2)}\ge \frac{1}{2n+2}$ And am now stuck. I just can't figure out how to show it's true.

Answer 1

You have it true for $n$ to prove for $n+1$ you need $\frac{1*3*5*...*(2n+1)}{2*4*....(2n+2)}\ge \frac{1}{2n+2}$ but by induction $\frac{1*3*5*....*(2n+1)}{2*4*...*(2n+2)}\ge \frac{1}{2n}\frac{2n+1}{2n+2}\ge \frac{1}{2n} \frac{2n}{2n+2}=\frac{1}{2n+2}$

Answer 2

Cancel out the $2n$ on both sides and obtain an equivalent inequality:

$ 3\cdot 5\cdots (2n-1) \geq 2\cdot 4\cdot 6\cdots (2n-2)$ which is clear since:

$3 \geq 2, 5 \geq 4,\cdots , 2n-1 \geq 2n-2$.