Mathematical induction problem with inequality

Question

I have the following problem where n is a positive integer $(n >= 1)$:

Prove that $\frac{1}{2n}\le\frac{1*3*5*...*(2n-1)}{2*4*...*2n}$

I know that I must start with the basic step showing that $P(1)$ is true as follows: $1/(2*1) = 1/2$ so $P(1)$ is true.

Now follows the induction step where I must show that "if $P(k)$ then $P(k+1)$" where $P(k)$ is $\frac{1}{2k}\le\frac{1*3*5*...*(2k-1)}{2*4*...*2k}$ and $P(k+1)$ is $\frac{1}{2(k+1)}\le\frac{1*3*5*...*(2k+1)}{2*4*...*2(k+1)}$

I will very much appreciate any help about how to continue.

Answer 1

You don't really need induction: your inequality is equivalent to $$ 1\leq\frac{3\times 5\times \cdots \times (2n-1)}{2\times 4\times \cdots\times (2n-2)}=\frac{3}{2}\times\frac{5}{4}\times\cdots\times\frac{2n-1}{2n-2}\cdot $$ The rightmost expression above is clearly no less than $1$.

Answer 2

$$\frac{1*3*5*...*(2k-1)*(2k+1)}{2*4*...*(2k)*(2k+2)} \geq \frac{1*(2k+1)}{(2k)*(2k+2)}$$ from the induction hypothesis.

The right side is greater than $\frac{1}{(2k+2)}$ since $\frac{2k+1}{2k} > 1$. Putting these inequalities together gives the desired result.