<s>[INST] To prove $1^1\cdot2^2\cdot 3^3...\cdot n^n<(\frac{2n+1}{3})^{\frac{n(n+1)}{2}} $ So we have to prove the following for $n\in N $ $$1^1\cdot 2^2\cdot 3^3...\cdot n^n<\left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}} $$
So I used concept of weighted means (arithmetic and geometric) used AM GM inequality.
$$AM=\frac{a_1w_1+a_2w_2+...+a_nw_n}{w_1+w_2+...+w_n}$$
$$GM=(a_1^{w_1}\cdot a_2^{w_2}\cdot...\cdot a_n^{w_n})^{\frac{1}{w_1+w_2+...+w_n}}$$
So here I let $w_1=1, w_2=2^1,w_3=3^1..$ and of course $a_1=1,a_2=2^1,a_3=3^2...$
So we get:
$$\frac{1^1+ 2^2+ 3^3...+ n^n}{\frac{n(n+1)}{2}}>(1^1\cdot 2^2\cdot 3^3...\cdot n^n)^{\frac{1}{\frac{n(n+1)}{2}}}$$
However on lhs, I cant deal with numerator, and I feel that if it can be simplified, I would get the answer. So please help or if possible suggest new method.
 [/INST] Use the AM-GM inequality [where there are $\frac{n(n+1)}{2}$ terms; indeed for each positive integer $i \le n$, there are $i$ terms of $i$]:
$$\frac{\sum_{1=1}^n i^2}{n(n+1)/2} \ \ge \ \sqrt[\frac{n(n+1)}{2}]{1^12^2 \cdots n^n},$$
where in the above inequality, the LHS represents the arithmetic mean of the above terms and the RHS the geometric mean of the above terms.
However, the equation
$$\frac{\sum_{1=1}^n i^2}{n(n+1)/2}  = \frac{(n)(n+1)(2n+1)}{6} \times \frac{2}{n(n+1)}$$
$$ = \frac{2n+1}{3} $$ also holds. So then combining this string of equations with the top AM-GM inequality, yields the inequality
$$\frac{2n+1}{3} \ \ge \ \sqrt[\frac{n(n+1)}{2}]{1^12^2 \cdots n^n} \ .$$
Raising each side of this to the $\frac{n(n+1)}{2}$-power yields the desired result.
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