All Questions
21
questions
4
votes
2
answers
637
views
Prove $\prod_{k=1}^n(1+a_k)\leq1+2\sum_{k=1}^n a_k$
I want to prove
$$\prod_{k=1}^n(1+a_k)\leq1+2\sum_{k=1}^n a_k$$
if $\sum_{k=1}^n a_k\leq1$ and $a_k\in[0,+\infty)$
I have no idea where to start, any advice would be greatly appreciated!
- 143
1
vote
3
answers
89
views
Prove that $\prod_{n \in \mathbb{N}}{(1-a_n)} \geq 1 - \sum_{n \in \mathbb{N}}{a_n}$
The following proof is obtained from this paper.
My question is how to obtain the inequality.
My guess is because of the following inequality:
$$\prod_{n \in \mathbb{N}}{(1-a_n)} \geq 1 - \sum_{n ...
- 15.9k
1
vote
1
answer
235
views
An inequality involving sums and products
I am curious to know whether the following holds or not.
If $n_1,n_2,n_3,m_1,m_2$ are positive integers strictly greater than 1 such that $$n_1+n_2+n_3 > m_1 +m_2$$
then $$n_1n_2n_3 \geq m_1m_2.$$
...
- 13
0
votes
1
answer
589
views
Generalized Holder Inequality
Let $a_i \in \mathbb R^n$ with $a_i = (a_{i}^j)_{j = 1 ... n} = (a_{i}^1, ... ,a_{i}^n)$ for $i = 1, ... , k$ and let $p_1,...,p_k \in \mathbb R_{>1}$ with $\frac1{p_1}+ ... + \frac1{p_k} = 1$
...
- 465
1
vote
1
answer
224
views
Prove $1 + \sum_{i=0}^n(\frac1{x_i}\prod_{j\neq i}(1+\frac1{x_j-x_i}))=\prod_{i=0}^n(1+\frac1{x_i})$
Prove the identity
$$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)$$
and hence deduce the inequality in Problem ...
- 4,996
12
votes
1
answer
885
views
Showing $\sum\limits^N_{n=1}\left(\prod\limits_{i=1}^n b_i \right)^\frac1{n}\le\sum\limits^N_{n=1}\left(\prod\limits_{i=1}^n a_i \right)^\frac1{n}$?
If $a_1\ge a_2 \ge a_3 \ldots $ and if $b_1,b_2,b_3\ldots$ is any rearrangement of the sequence $a_1,a_2,a_3\ldots$ then for each $N=1,2,3\ldots$ one has
$$\sum^N_{n=1}\left(\prod_{i=1}^n b_i \right)^...
- 2,048