<s>[INST] Inequality with induction. I was wondering if my solution to this problem is correct:
Show that for every integer $n\ge1$ the following is true:
$\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\le\frac{1}{\sqrt{3n+1}}$
Base case:
$\frac{2\cdot1-1}{2}\le\frac{1}{\sqrt{3*1+1}}$
$\frac{1}{2}\le\frac{1}{2}$
$\Rightarrow \text{which is true.}$
Induction assumption:
$\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\le\frac{1}{\sqrt{3n+1}}$
Which means we want to prove the following:
$\prod\limits_{k=1}^{n+1}\frac{2k-1}{2k}\le\frac{1}{\sqrt{3(n+1)+1}}$
$\prod\limits_{k=1}^{n+1}\frac{2k-1}{2k}\le\frac{1}{\sqrt{3n+4}}$
Which is the same as:
$\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\cdot(\frac{2(n+1)-1}{2(n+1)})\le\frac{1}{\sqrt{3n+4}}$
$\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\cdot(\frac{2n+1}{2n+2)})\le\frac{1}{\sqrt{3n+4}}$
And then, we know the following is true:
$\frac{1}{\sqrt{3n+4}}\lt\frac{1}{\sqrt{3n+1}}$
Which lets us say the following:
$\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\cdot(\frac{2n+1}{2n+2)})\le\frac{1}{\sqrt{3n+4}}\le\frac{1}{\sqrt{3n+1}}$
The beforementioned statement is true because $2n+2\gt2n+1$ and $n\ge1$, which means that:
$0\lt\frac{2n+1}{2n+2}\lt1$
We can now that this is true according to the induction assumption. Which proves the original statement.
 [/INST] To conclude for the induction step we should use the induction hypothesis that is
$$\prod\limits_{k=1}^{n+1}\frac{2k-1}{2k}=\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\cdot\left(\frac{2n+1}{2n+2}\right)\stackrel{Ind.Hyp.}\le\frac{1}{\sqrt{3n+1}}\cdot\left(\frac{2n+1}{2n+2}\right)\stackrel{?}\le \frac{1}{\sqrt{3(n+1)+1}}$$
and the latter is true indeed
$$\frac{1}{\sqrt{3n+1}}\cdot\left(\frac{2n+1}{2n+2}\right)\stackrel{?}\le \frac{1}{\sqrt{3(n+1)+1}}$$
$$\iff \left(\frac{2n+1}{2n+2}\right)^2\le \frac{3n+1}{3n+4}$$
$$\iff (4n^2+4n+1)(3n+4)\le (4n^2+8n+4)(3n+1)$$
$$\iff 12n^3+28n^2+19n+4\le 12n^2+28n^2+20n+4$$
$$\iff 19\le 20$$
</s>