4
votes
2answers
94 views

inequality $\prod\limits_{k=1}^n\frac{2k-1}{2k}\lt\frac1{\sqrt{3n}}$

$n$ is a positive integer, then $$\prod_{k=1}^n\frac{2k-1}{2k}\lt\frac1{\sqrt{3n}}$$ with mathematical induction, we can prove this. But I would love to find a wonderful method without ...
0
votes
0answers
27 views

How to derive this inequality containing power series? (equations are contained in the body) (Changed)

As I read a paper, I don't know how do I derive inequality, $$\frac{\prod^N_{i=1}4^{b_i/N}(1-4^{-b_i})^{1/N}}{12}\ge \frac{4^{R/N}}{16} \\ \frac{\prod^N_{i=1}(4^{b_i}-1)^{1/N}}{12}\ge ...
2
votes
2answers
85 views

The floor of a product of fractions

Evaluate: $ \displaystyle \Bigg \lfloor \prod_{n=0}^{248} \frac{33+8n}{29+8n} \Bigg \rfloor= \Bigg \lfloor \frac{33}{29} \times \frac{41}{37} \times \frac{49}{45} \times\ ...\ \times ...
2
votes
1answer
83 views

Is anyway to prove this: $\prod_{k=1}^{n}(a_{k})< (1/n^n)*(\sum_{k=1}^{n}(\sqrt{1+a_{k}*a_{k+1}}))^n$

$$ \prod\limits_{k=1}^{n}a_{k} < {1 \over n^{n}}\left(\,\sum_{k = 1}^{n}\,\sqrt{1+a_{k}\,a_{k+1}\,}\,\right)^n $$ ak and n are positive real number greater than 0. EDIT: a_{k+1} becomes a_{1} ...
1
vote
1answer
65 views

Prove $1 + \sum_{i=0}^n(\frac1{x_i}\prod_{j\neq i}(1+\frac1{x_j-x_i}))=\prod_{i=0}^n(1+\frac1{x_i})$

Prove the identity $$1 + \sum_{i=0}^n \left(\frac1{x_i}\prod_{j\neq i} \left(1+\frac1{x_j-x_i} \right) \right)=\prod_{i=0}^n \left(1+\frac1{x_i} \right)$$ and hence deduce the inequality in Problem ...
5
votes
2answers
875 views

Proving the AM:GM inequality

I am doing past exam papers preparing for the finals and I came across this questions about three times: Prove that: $$\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\geq \sqrt[n]{a_{1}.a_{2}...a_{n}}$$ ...
0
votes
2answers
34 views

An inequality involving a product

Let $x_1\in(0,1)$ and $a_1,\ldots,a_n\ge-1$ reals. We know that \begin{equation} \prod_{i=1}^n (1+x_1a_i) < 1 \end{equation} Does it then also hold true that \begin{equation} \prod_{i=1}^n ...
5
votes
1answer
278 views

Showing $\sum\limits^N_{n=1}\left(\prod\limits_{i=1}^n b_i \right)^\frac1{n}\le\sum\limits^N_{n=1}\left(\prod\limits_{i=1}^n a_i \right)^\frac1{n}$?

If $a_1\ge a_2 \ge a_3 \ldots $ and if $b_1,b_2,b_3\ldots$ is any rearrangement of the sequence $a_1,a_2,a_3\ldots$ then for each $N=1,2,3\ldots$ one has $$\sum^N_{n=1}\left(\prod_{i=1}^n b_i ...