<s>[INST] Proof by induction $1\cdot 3 \cdot 5\cdots (2n-1) / (2 \cdot 4\cdot 6\cdots 2n) \leq 1/ \sqrt{n+1}, \forall n \geq 1, n\in \mathbb N$ Prove by induction for
$$ \frac{\left(1\right)\left(3\right)\left(5\right)...\left(2n-1\right)}{\left(2\right)\left(4\right)\left(6\right)...\left(2n\right)}\le\frac{1}{\sqrt{n+1}}\quad \text{for all integers}\quad n \ge 1$$
My current inductive step:
$$ \frac{(2(k+1))!}{2^{2\left(k+1\right)}((k+1)!)^2} = \frac{2k+1}{2(k+1)} \cdot\frac{(2k)!}{2^{2k}\left(k!\right)^2}$$
I'm not sure how to proceed to show 
$$ \frac{2k+1}{2(k+1)} \cdot \frac{(2k)!}{2^{2k}\left(k!\right)^2} \le \frac{1}{\sqrt{k+2}}$$
 [/INST] There's no need to use factorials here, you can do it with basic arithmetic. The base case is
$$\frac{1}{2} \leq \frac{1}{\sqrt{2}},$$
which you can square both sides to find that it's true.
Then we note that if $f(n)$ is the fraction on the left of the inequality, then $f(n+1) = f(n)\frac{2n+1}{2n+2}$.
So we assume for induction that $f(n) \leq \frac{1}{\sqrt{n+1}}$, then:
$$f(n) \leq \frac{1}{\sqrt{n+1}}$$
$$f(n)\frac{2n+1}{2n+2} \leq \frac{1}{\sqrt{n+1}}\frac{2n+1}{2n+2}$$
$$f(n+1) \leq \frac{2n+1}{\sqrt{n+1}(2n+2)}$$
Now we want to know if
$$\frac{2n+1}{\sqrt{n+1}(2n+2)} \leq \frac{1}{\sqrt{n+2}}$$
$$(2n+1)\sqrt{n+2} \leq \sqrt{n+1}(2n+2)$$
$$(2n+1)^2(n+2) \leq (n+1)(2n+2)^2$$
$$4 n^3 + 12 n^2 + 9 n + 2 \leq 4 n^3 + 12 n^2 + 12 n + 4$$
$$9 n \leq 12 n + 2$$
Which is easy to see to be true for any positive $n$.
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