<s>[INST] Prove that $1\cdot \frac{1}{2^2}\cdot ...\cdot \frac{1}{n^n}< \left(\frac{2}{n+1}\right)^\frac{n(n+1)}{2}$ Prove that $$1\cdot \frac{1}{2^2}\cdot ...\cdot \frac{1}{n^n}< \left(\frac{2}{n+1}\right)^\frac{n(n+1)}{2}$$ where $n$ is a positive integer.
My book suggests using AM-GM, but I couldn't do it. I just applied AM-GM to the numbers in the LHS, but it looks like I need some more upper bounds.
 [/INST] By AM-GM, we have: 
$ \frac{1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+...+(\frac{1}{n}+...+\frac{1}{n})}{\frac{n(n+1)}{2}}$
$=\frac{1+1+1+...+1}{\frac{n(n+1)}{2}}$, where there are $n$ copies of $1$.
$=\frac{n}{\frac{n(n+1)}{2}}=\frac{2}{n+1}$
$\geq \sqrt[\leftroot{-2}\uproot{2}\frac{n(n+1)}{2}]{1 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot ... \cdot (\frac{1}{n}...\frac{1}{n})}$
$=\sqrt[\leftroot{-2}\uproot{2}\frac{n(n+1)}{2}]{1 \cdot \frac{1}{2^2} \cdot ... \cdot \frac{1}{n^n}}$
But $\frac{2}{n+1} \geq \sqrt[\leftroot{-2}\uproot{2}\frac{n(n+1)}{2}]{1 \cdot \frac{1}{2^2} \cdot ... \cdot \frac{1}{n^n}} \Rightarrow 1 \cdot \frac{1}{2^2} \cdot ... \cdot \frac{1}{n^n} \leq (\frac{2}{n+1})^{\frac{n(n+1)}{2}} $, thus concluding our proof.
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