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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.

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    $\begingroup$ More precisely, we should have $\geq$ rather than $>$, since we clearly have equality of both LHS and RHS when $n=1$. $\endgroup$
    – Hello
    Commented Apr 22, 2020 at 16:00
  • $\begingroup$ @ ChemistryGeek does my solution make sense to you? $\endgroup$
    – Hello
    Commented Apr 22, 2020 at 16:15
  • $\begingroup$ @ONGSEEHAI yes, it's fine, thanks $\endgroup$ Commented Apr 22, 2020 at 21:51

3 Answers 3

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Apply AM/GM to the sequence $a_1,\ldots,a_{n(n+1)/2}$ which looks lie $1,1/2,1/2,1/3,1/3,1/3,\ldots,1/n$ ($1/k$ occurs $k$ times for $1\le k\le n$). The $AM$ is $2/(n+1)$ etc.

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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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  • $\begingroup$ Note that the main trick behind applying AM-GM is to decompose and view each term of the form $\frac{1}{k^k}$ as $k$ equivalent copies of $\frac{1}{k}$. $\endgroup$
    – Hello
    Commented Apr 22, 2020 at 16:04
  • $\begingroup$ Note that there are $\frac{n(n+1)}{2}$ terms in our first line. To understand why, observe that there is $1$ copy of $1$, $2$ copies of $2$, ... , $n$ copies of $n$. And we thus have our standard triangular sum of the first $n$ natural numbers. $\endgroup$
    – Hello
    Commented Apr 22, 2020 at 16:08
  • $\begingroup$ Would appreciate if anyone could provide suggestions (if any) on how to improve my presentation. Thanks in advance! $\endgroup$
    – Hello
    Commented Apr 22, 2020 at 16:10
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Also induction works well here.

$$1\cdot \dfrac{1}{2^2}\cdots \dfrac{1}{n^n}\cdot\dfrac{1}{(n+1)^{n+1}} \le \left(\dfrac{2}{n+1}\right)^\frac{n(n+1)}{2}\cdot \dfrac{1}{(n+1)^{n+1}} \\= \Big[ \Big(\dfrac{2}{n+1}\Big)^{\frac{n}{2}}\cdot \dfrac{1}{n+1} \Big]^{n+1} \le(?)\ \left(\dfrac{2}{n+2}\right)^\frac{(n+1)(n+2)}{2}=\Big[ \Big(\dfrac{2}{n+2}\Big)^{\frac{n+2}{2}} \Big]^{n+1}.$$

To conclude you have to show that

$$\Big(\dfrac{2}{n+1}\Big)^{\frac{n}{2}} \cdot \dfrac{1}{n+1} \le \Big(\dfrac{2}{n+2}\Big)^{\frac{n+2}{2}}=\Big(\dfrac{2}{n+2}\Big)^{\frac{n}{2}}\cdot \dfrac{2}{n+2}.$$

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