2

If $ n $ be a positive integer $>1$, prove that $$2^{n(n+1)}\gt(n+1)^{n+1}\biggl(\frac{n}{1}\biggr)^{n}\biggl(\frac{n-1}{2}\biggr)^{n-1}...\biggl(\frac{2}{n-1}\biggr)^{2}\biggl(\frac{1}{n}\biggr)$$

Math-Boy
  • 91
  • 3
    Hello. You have a track record of posting questions with no indication of what you have tried, or of the source of the problem. Please give your question some context. http://meta.math.stackexchange.com/questions/9959/how-to-ask-a-good-question/9960#9960 – not all wrong Dec 10 '15 at 13:50

2 Answers2

3

Consider $(n+1)$ positive numbers $\binom{n}{0}, \binom{n}{1}, ...\binom{n}{n}$. $$ A.M.= \frac{\binom{n}{0}+ \binom{n}{1}+ ...+\binom{n}{n}}{n+1}=\frac{(1+1)^n}{n+1}=\frac{2^n}{n+1}$$ $$G.M.=\sqrt[(n+1)]{\binom{n}{0} \binom{n}{1}. ...\binom{n}{n}}=\sqrt[(n+1)]{\biggl(\frac{n}{1}\biggr)^n \biggl(\frac{n-1}{2}\biggr)^{n-1} \biggl(\frac{n-2}{3}\biggr)^{n-2}...\biggl(\frac{2}{n-1}\biggr)^{2} \biggl(\frac{1}{n}\biggr)^{1}} $$ Applying A.M.> G.M., we get $$ \biggl(\frac{2^n}{n+1}\biggr)^{n+1}> \biggl(\frac{n}{1}\biggr)^n \biggl(\frac{n-1}{2}\biggr)^{n-1} \biggl(\frac{n-2}{3}\biggr)^{n-2}...\biggl(\frac{2}{n-1}\biggr)^{2} \biggl(\frac{1}{n}\biggr)^{1}$$

Math-Boy
  • 91
1

This is too long to be a comment:You can simplify the RHS to be one of the following based on the parity of $n$. $$(n+1)^{(n+1)}\prod_{i=1}^{\frac{n-1}{2}}\left(\frac{n-i}{i}\right)^{2i+1}\quad\text{or}\quad (n+1)^{(n+1)}\left(\frac{n}{2}\right)^\frac{n}{2}\prod_{i=1}^{\frac{n-2}{2}}\left(\frac{n-i}{i}\right)^{2i+1}$$ clearly the one on the right is much bigger, and so you should work from there, but it's not at all clear to me that that one actually satisfies this inequality.

Stella Biderman
  • 31,155