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.

This is not a proof since working for large values of $n$
$$\prod_{k=1}^n k^k=H(n)$$ where $H(n)$ is the hyperfactorial function.
Expanding its logarithm
$$\log (H(n))=-\frac 14 n^2+\frac 1{12} \left(6 n^2+6 n+1\right)\log(n)+\log (A)+\frac{1}{720 n^2}\left(1-\frac{1}{7 n^2}+\frac{1}{14
   n^4}+O\left(\frac{1}{n^6}\right) \right)$$
Doing the same for the logarithm of the rhs
$$\log\left(\frac{\text{rhs}}{\text{lhs}}\right)=\log \left(\frac{4 e}{9}\right)\,\frac{n(n+1)}4-\frac 1{12}\log(n)+\left(\frac{3}{16}-\log (A)\right)+O\left(\frac{1}{n}\right)$$
