All Questions
21
questions
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
$$\sum^N_{n=1}\left(\prod_{i=1}^n b_i \right)^...
- 2,048
6
votes
0
answers
97
views
Prove that if $x_{1}x_{2}...x_{n}=1$ then $\frac{1}{1+x_{1}+x_{1}x_{2}}+...+\frac{1}{1+x_{n-1}+x_{n-1}x_{n}}+\frac{1}{1 +x_{n}+x_{n}x_{1}}\ge 1$ [duplicate]
Prove that if $x_{1}x_{2}...x_{n}=1$ then $\frac{1}{1+x_{1}+x_{1}x_{2}}+...+\frac{1}{1+x_{n-1}+x_{n-1}x_{n}}+\frac{1}{1 +x_{n}+x_{n}x_{1}}\ge 1$.
$x_{1},x_{2},...,x_{n}$ are positive real numbers, and ...
- 288
5
votes
3
answers
195
views
Prove that $\frac{\sqrt[n]{\prod_{k = 1}^nx_n}}{m} \ge n - 1$ where $\sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m}$.
Given positives $x_1, x_2, \cdots, x_{n - 1}, x_n$ such that $$\large \sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m}$$. Prove that $$\large \frac{\displaystyle \sqrt[n]{\prod_{k = 1}^nx_n}}{m} \ge n - ...
- 4,246
4
votes
2
answers
272
views
Sum of positive elements divided by their "weighted" product - inequality
I have following expression,
$$ \frac{\sum_{i=1}^n x_i}{\prod_{i=1}^nx_i^{p_i}} $$
where $p_i$s satisfy $\sum p_i = 1$ and $p_i \in [0,1]$ and $x_i\geq0$, $\forall i \in 1\dots n$.
I think that ...
- 221
4
votes
3
answers
104
views
$Q\le \prod \frac{5+2x}{1+x}\le P$ find $P,Q$
if $x,y,z,$ are positives and $x+y+z=1$ and $$Q\le \prod_{cyc} \frac{5+2x}{1+x}\le P$$ find maximum value of $Q$ and minimum value of $P$
This is actually a question made up myself ,so i don,t know ...
- 11.9k
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
4
votes
1
answer
3k
views
Proof of Inequality involving sum and product without induction
How could you prove these inequalities wihout induction:($a_k$ are non-negative)
1)$\prod_{k=1}^n(1+a_k)\ge1+\sum_{k=1}^n a_k$
2)$\prod_{k=1}^n(1+a_k)\le1+\frac{\sum_{k=1}^na_k}{1!}+\ldots+\frac{(\...
- 7,085
3
votes
2
answers
178
views
Prove that $\frac1{a(1+b)}+\frac1{b(1+c)}+\frac1{c(1+a)}\ge\frac3{1+abc}$
I tried doing it with CS-Engel to get $$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c}
$$
I thought that maybe proof that $$
\frac{1}{a+b+c+a b+b c+a c} \geq \...
3
votes
2
answers
81
views
$AM-GM$-ish inequality
Suppose $x_0, \cdots, x_n$ are positive reals. Suppose that:
$$\sum_{i = 0}^n \frac{1}{1+x_i} \leq 1$$
Then show that:
$$\prod_{i=0}^{n} x_i \geq n^{n+1} $$
I got to this problem by rewriting problem ...
- 4,284
2
votes
1
answer
84
views
Proving an inequality consisting of sums and products
I have a tricky inequality (related to some previous ones that I posted) which I am yet again stuck trying to solve. I have confirmed that it is true through simulation (at least up until overflow ...
- 183
2
votes
0
answers
64
views
How to prove the following inequality with complete induction?
Let $n \in \mathbb{N}$, and let $a_1, ... , a_n > 0.$
Show that:
I got the hint that we have to use this induction step for the induction proof:
And thats what I got so far in the Induction step ...
- 33
1
vote
3
answers
89
views
Prove that $\prod_{n \in \mathbb{N}}{(1-a_n)} \geq 1 - \sum_{n \in \mathbb{N}}{a_n}$
The following proof is obtained from this paper.
My question is how to obtain the inequality.
My guess is because of the following inequality:
$$\prod_{n \in \mathbb{N}}{(1-a_n)} \geq 1 - \sum_{n ...
- 15.9k
1
vote
1
answer
63
views
Prove $x^{y+1}z+y^{z+1}x+z^{x+1}y\geq x^2y^2z^2$
Let $x>1$, $y>1$ and $z>1$ be such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. Prove that:
$$x^{y+1}z+y^{z+1}x+z^{x+1}y\geq x^2y^2z^2.$$
When $x=y=z=3$, both sides are equal to $3^6$. ...
- 7,194
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
1
vote
2
answers
125
views
Mathematicals inequalities
For $$x,y,z>0 $$
Prove that
$$(2xyz)^2 \ge (x^3+y^3+z^3+xyz)(x+y-z)(y+z-x)(z+x-y)$$
I have tried a famous inequalities: $$(x+y-z)(y+z-x)(z+x-y) \le xyz$$
So the problem is:
$$3xyz \ge x^3+y^3+z^3$$
...
- 21