A problem of AM>GM

Question

If $n$ is a positive integer with $n> 1$, prove that $$2^{n(n+1)}>(n+1)^{(n+1)}\cdot\left(\frac{n}{1}\right)^n\cdot\left(\frac{n-1}{2}\right)^{(n-1)}\cdots\left(\frac{2}{n-1}\right)^2\cdot\frac{1}{n}$$

For solving it I have considered the numbers $\displaystyle{n+1,\,n,\,\frac{n-1}{2},\,\ldots,\frac{2}{n-1},\,\frac{1}{n}}$ with associated weights $n+1,n,n-1,\ldots,2,1$ and applied AM>GM $$\left[{(n+1)^2+n^2+\frac{(n+1)^2}{2}+\ldots+\frac{2^2}{(n-1)}+\frac{1}{n}}\over{\frac{(n+1)\cdot(n+2)}{2}}\right]^{\tfrac{(n+1)\cdot(n+2)}{2}}$$

but cannot solve it please help

Hello, welcome to Math.SE. For some basic information about writing maths at this site see e.g. here, here, here and here. — Lord_Farin, Jun 28 '13 at 09:44
OP: Why did you delete the part of the question showing the computations you tried to solve it? — Did, Jun 28 '13 at 10:49
this is a problem from the inequality chapter and has come in CU 2011 — kinkar, Jun 28 '13 at 10:49
Oftopic, since somebody had done it for you. Again: why the deletion? — Did, Jun 28 '13 at 10:58
How so? The formula they LaTeXed and the one you had typed are identical. — Did, Jun 28 '13 at 11:04
the power of the expression was written for the denominator which I wrote was for the whole of it — kinkar, Jun 28 '13 at 11:07
As I said, the formula they LaTeXed and the one you had typed are identical. — Did, Jun 28 '13 at 11:50
I proposed the exact same edit, yet it was rejected... Anyway, instead of removing, why didn't you just post an image of the equation if you didn't know how to LaTeX — Abhimanyu Pallavi Sudhir, Jun 29 '13 at 04:27

score 3 · Answer 1 · answered Jun 29 '13 at 06:40

Hint:

$$\left(\frac{n}{1}\right)^n\cdot\left(\frac{n-1}{2}\right)^{(n-1)}\cdots\left(\frac{2}{n-1}\right)^2\cdot\frac{1}{n} = \binom{n}{0} \binom{n}{1} \cdots \binom{n}{n}$$ where $\binom{n}{k}$ is the usual binomial coefficient.

A problem of AM>GM

1 Answers1