Prove by induction that $\frac{1}{2n}\leq\frac{1\text{·}3\cdot5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}$ What would be the right way to solve this by induction proof?
$$\frac{1}{2n}\leq\frac{1\text{·}3\cdot5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}$$
This is what I've done (reference https://www.slader.com/discussion/question/prove-that-12n-1-3-5-2n-12-4-2n-whenever-n-is-a-positive-integer/#):

*

*Show that $S\left(n+1\right)$ by induction proof. This is
$$\frac{1}{2(n+1)}\leq\frac{1\text{·}3\text{·}5\text{·}\ldots\text{·}(2n+1)}{2+4+6+\ldots+2(n+2)}$$
Multiplying both sides of the equation $\frac{2n\text{·}(2(n+1)-1)}{2n\text{·}(2(n+1)-1)}=\frac{2n\text{·}(2n+1)}{2n\text{·}(2n+1)}$
$$\frac{1}{2n}\leq\frac{1\text{·}3\text{·}5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}\times\frac{2n\text{·}(2(n+1)-1}{2n\text{·}(2(n+1)-1}$$
$$\frac{1}{2n}\times\frac{2n\text{·}(2n+1)}{2n\text{·}(2n+1)}$$
Rewriting we have the following
\begin{array}{c}
\frac{2n+1}{2n(2n+1)}\\
\frac{1}{2n+1}+\frac{1}{2n(2n+1)}
\end{array}
$$\frac{1}{2n}\leq\frac{1}{2n+1}+\frac{1}{2n(2n+1)}$$

Let $\varphi, \psi:\mathbb N\to\mathbb N$ given by
\begin{aligned}
\varphi(n) &= \sum_{k=1}^n 2k = n(n+1)\\
\psi(n) &= \prod_{k=1}^n (2k-1)\\
\end{aligned}
and notice that
\begin{aligned}
\varphi(n+1) &= \frac{n+2}n\varphi(n)\\
\psi(n+1) &= (2n+1)\psi(n)\\
\end{aligned}
If you assume that, for a certain $n\in\mathbb N$, the following is true
$$\frac 1{2n}\le \frac{1\times 3\times\ldots\times (2n-1)}{2 + 4 + \ldots + 2n} = \frac{\psi(n)}{\varphi(n)},$$
then you may write
\begin{aligned}
\frac 1{2(n+1)} 
& = \frac{2n}{2(n+1)}\frac1{2n}\\
& \le \frac{2n}{2(n+1)}\frac{\psi(n)}{\varphi(n)}\\
& = \frac{2n}{2(n+1)}\frac1{2n+1}\frac{n+2}{n}\frac{\psi(n+1)}{\varphi(n+1)}\\
& = \frac{n+2}{(n+1)(2n+1)}\frac{\psi(n+1)}{\varphi(n+1)}\\
& \le \frac{\psi(n+1)}{\varphi(n+1)},
\end{aligned}
where the last inequality follows from the fact that
$$
\frac{n+2}{(n+1)(2n+1)} = \frac{1}{2n+1}\left(1 + \frac1{n+1}\right) \le \frac 12\left(1 + \frac 12\right) = 1.
$$
