<s>[INST] Show that $\frac{\left(2n\right)!}{2^{n}n!} \ge \frac{\left(2n\right)!}{\left(n+1\right)^{n}}$. I am attempting to solve this by induction, yet I am stuck on the last part. Below is my attempted solution.
Show that $\frac{\left(2n\right)!}{2^{n}n!} \ge \frac{\left(2n\right)!}{\left(n+1\right)^{n}}, \forall n \in \mathbf{N} $.
My attempted solution, by induction:
For $n=1$ we have,
$\frac{\left(2\right)!}{2^{1} \cdot 1!} \ge \frac{\left(2\right)!}{\left(1+1\right)^{1}} $
$2 \ge 1$.
Thus, the inequality holds for $n=1$.
Now assume that it holds $\forall k \in \mathbf{N}$.
Then, for $k+1$,
$\left(k+1\right) \frac{\left(2k\right)!}{2^{k}k!} \ge \frac{\left(2k\right)!}{\left(k+1\right)^{k}} \left(k+1\right) $.
This is where I am stuck.
 [/INST] By the AM-GM inequality you have
$$\sqrt{1 \cdot n} \leq \frac{n+1}{2} \\
\sqrt{2 \cdot (n-1)} \leq\frac{n+1}{2} \\
...\\
\sqrt{n \cdot 1} \leq \frac{n+1}{2} $$
Multiplying everything together you get
$$n! \leq \left( \frac{n+1}{2}\right)^n$$
</s>