Verify an inequality using induction: $\frac1{2n}\le \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}$

Question

How can I verify this inequality by using induction?

$$\frac1{2n}\le \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}$$

I understand the principle of induction, but I'm struggling with how my teacher is solving this problem. I Highlighted the area I'm stuck at In the solution – why are we able to take $1/2k$ and see if its less than the left side of our inducted hypothesis? (view the highlighted part of the photo)

Any suggestions would be great – I really don't understand how to solve this problem. If I could get a step by step walk through of this proof with side notes, I'd be very grateful

Answer 1

In this screenshot from your image call the LHS and RHS of the first inequality (which is a consequence of the induction hypothesis) $B$ and $C$ respectively. The first inequality is stating that $\boxed{B\le C}$.

Call the LHS of the second inequality, marked in yellow by you, to be $A$.

You want to prove $A\le C$, and you have already proved $B\le C$ (mentioned in the first block above), so
if you can prove $A\le B$, then you can combine the consequence of the first inequality, i.e $\boxed{B\le C}$ with $A\le B$ to get $$A\le B \le C \implies A \le C$$ which would prove the claim. So we want to prove (as written by your teacher) that $A\le B$.