# 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.

• Take the log of both sides. You'll find a familiar sum on the left and a familiar product on the right. Dec 14, 2022 at 17:04
• @KshitijKumar You shouldn't. Take the log of both sides and turn the left into a sum and the right into a simple product. Once you get the sum it almost falls into your lap. Dec 14, 2022 at 17:09
• Your stuff is right except a typo. Your LHS should be $\left( 1\cdot1+2\cdot 2 +\cdots + n\cdot n\right)/\text{number of terms} = \frac{2n+1}{3}$. Also AM is greater than, not less than the GM. Then raise the power of both sides. Dec 14, 2022 at 18:40
• Why were all the answers below deleted?
– Mike
Dec 14, 2022 at 19:28
• @user2661923 That deleted answer mixed up which of $\log n$ and $\log(\frac{2n+1}{3})$ was bigger. Despite a few users saying it does, I have yet to see how taking logs of both sides yields a proof.
– anon
Dec 14, 2022 at 20:21

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.

AMGM is the right idea, you just applied it wrong. As you say, the inequality is

$$\frac{w_1x_1+\cdots+w_nx_n}{w_1+\cdots+w_n}\ge(x_1^{w_1}\cdots x_n^{w_n})^{1/(w_1+\cdots+w_n)}.$$

(If we define $$p_k=w_k/(w_1+\cdots+w_n)$$, this reads $$p_1x_1+\cdots+p_nx_n\ge x_1^{p_1}\cdots x_n^{p_n}$$.)

In your case, if you define $$w_k=x_k=k$$ for $$k=1,\cdots,n$$ the inquality becomes

$$\frac{1^2+\cdots+n^2}{1+\cdots+n}\ge(1^1\cdots n^n)^{1/(1+\cdots+n)}.$$

Using $$1+\cdots+n=\frac{n(n+1)}{2}$$ and $$1^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$$ you should be able to finish.

(After writing this I read up and apparently Giant Ray pointed this out in the comments.)

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 $$\log\left(\frac{\text{rhs}}{\text{lhs}}\right)=\log \left(\frac{4 e}{9}\right)\,\frac{n(n+1)}4-\frac 1{12}\log(n)+\left(\frac{3}{16}-\log (A)\right)+O\left(\frac{1}{n}\right)$$