I thought I was doing really well and then I spent 90 minutes confused and I am not sure what the next step is.
I am supposed to show:
$\prod_{k=1}^{n} \frac{(2k-1)}{2k} \leq \frac{1}{\sqrt{3n+1}}$
So I did the base case:
$\prod_{k=1}^{1} \frac{((2*1)-1)}{(2*1)} \leq \frac{1}{\sqrt{(3*1)+1}})=\frac{1}{2}\leq \frac{1}{2}$
Then I got half way through the induction step:
$\prod_{k=1}^{n+1} \frac{(2k-1)}{2k} = \prod_{k=1}^{n} \frac{(2k-1)}{2k} * \frac{(2(n+1)-1)}{2(n+1)}$
So I was okay with translating the product operator for k+1 into familiar terms.
$\leq \frac{1}{\sqrt{3n+1}}*\frac{(2n+1)}{(2n+2)}$
Then, since $\frac{(2n+1)}{(2n+2)}$ must be positive, when I multiply it with the RHS, the inequality will hold.
Now I think I have to start playing with it, until I get$\frac{1}{\sqrt{3(n+1)+1}}$, but all of my attempts seemed rather inelegant. I figured I could just multiply by the reciprocal of $\frac{(2n+1)}{(2n+2)}$, since it's greater than $\frac{(2n+1)}{(2n+2)}$, which would also not change the inequality, but then I'd just have $\frac{1}{\sqrt{3n+1}}$....and I still wouldn't know how to get it to $\frac{1}{\sqrt{3(n+1)+1}}$. While it seems logically clear, that the square root of a larger number yields a larger number, I don't know how to write it in math language. Can someone help?
What you should do is take the square of the both sides of the hint. Since they are positive numbers the direction of inequality won't change then take the power $-1$ of both sides and when you perform the division on the left side you will see that there is an additional term besides $(3n+4)$ which makes it bigger than $(3n+4)$
$$\begin{align} \left(\frac{2n+1}{2n+2}\frac1{\sqrt{3n+1}}\right)^2&\le\left(\frac1{\sqrt{3n+4}}\right)^2\\[5pt] \frac{(2n+1)^2}{(2n+2)^2(3n+1)}&\le\frac1{3n+4}\\[5pt] \frac{(2n+2)^2(3n+1)}{(2n+1)^2}&\ge3n+4\\[5pt] \frac{(4n^2+8n+4)(3n+1)}{4n^2+4n+1}&\ge3n+4\\[5pt] \frac{12n^3+28n^2+20n+4}{4n^2+4n+1}&\ge3n+4\\[5pt] \boxed{3n+4}+\frac n{4n^2+4n+1}&\ge\boxed{3n+4} \end{align}$$