All Questions
46
questions
0
votes
1
answer
54
views
Why is $\sum_{m=0}^{\lfloor xs\rfloor} 2 \binom{s}{m} p^m (1-p)^{s-m} \leq 2\exp{\left(-\frac{2(\lfloor xs\rfloor - sp)^2}{s}\right)}$
I am trying to understand few of the mathematical steps I have encountered in a paper, there are two of them
(a) $\sum_{m=0}^{\lfloor xs\rfloor} 2 \binom{s}{m} p^m (1-p)^{s-m} \leq 2\exp{\left(-\frac{...
- 141
0
votes
0
answers
44
views
Inequality with Products and Sums
I need help to find a proof for the following inquality.
Assuming that $ 0 \leq c_i \leq 1 $ and $ 0 \leq d_i \leq 1 $, show that
$$
\prod_{i=1}^N (c_i + d_i - c_i d_i) \geq \prod_{i=1}^N c_i + \prod_{...
- 778
0
votes
0
answers
98
views
If $\sum_{i=1}^n x_i \ge a$, then what can we know about $\sum_{i=1}^n \frac{1}{x_i}$?
Suppose that $$\sum_{i=1}^n x_i \ge a$$
where $a>0$ and $x_i\in (0, b]$ for all $i$. Are there any bounding inequalities we can determine for $$\sum_{i=1}^n \frac{1}{x_i}?$$
I understand that $\...
- 1,615
0
votes
2
answers
92
views
Can someone give me a hint to this question concerning $\sum_{i=1}^n |x-i|$?
Find the smallest positive integer $n$ for which
$|x − 1| + |x − 2| + |x − 3| + · · · + |x − n| \geq 2022$
for all real numbers $x$.
I don't think I can combine any of these terms, right? So I started ...
- 13
3
votes
3
answers
167
views
find a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}$
Let $\{x\} = x-\lfloor x\rfloor$ be the fractional part of $x$. Order the (real) solutions to $\sqrt{\lfloor x\rfloor \lfloor x^3\rfloor} + \sqrt{\{x\}\{x^3\}} = x^2$ with $x\ge 1$ from smallest to ...
- 2,031
7
votes
1
answer
183
views
Show that $|x_{k+1}-x_k| \leq 1$ (for $0<k<n$) implies $\sum_{k=1}^n |x_k| - \left|\sum_{k=1}^n x_k\right|\leq\lceil(n^2-1)/4\rceil$.
Let $n\ge 1$ be a positive integer and let $x_1,\cdots, x_n$ be real numbers so that $|x_{k+1}-x_k|\leq 1$ for $k=1,2,\cdots, n-1$. Show $$\sum_{k=1}^n |x_k| - \left|\sum_{k=1}^n x_k\right|\leq \left\...
- 1,837
1
vote
1
answer
64
views
If $a_sk^s+a_{s-1}k^{s-1}+...+a_0$ is the basis representation of $n$ with respect to the basis $k$. Then, $0<n\leq k^{s+1}-1$.
If $a_sk^s+a_{s-1}k^{s-1}+...+a_0$ is the basis representation of $n$ with respect to the basis $k$. Then, $$0<n\leq k^{s+1}-1$$.
My attempt:-
By basis represantation, we know that $0\leq a_j<k,...
- 847
1
vote
1
answer
95
views
Prove that $\frac{1}{kn} + \frac{1}{kn + 1} + \dotsb + \frac{1}{kn + n - 1} > n \left(\sqrt[n]{\frac{k+1}{k}} - 1 \right)$
Let $k,n \in \mathbb{Z}^+$ with $n > 1$. Prove that $$\frac{1}{kn} + \frac{1}{kn + 1} + \dotsb + \frac{1}{kn + n - 1} > n \left(\sqrt[n]{\frac{k+1}{k}} - 1 \right)$$
I roughly observe that AM-...
- 374
15
votes
2
answers
607
views
The inequality $\,2+\sqrt{\frac p2}\leq\sum\limits_\text{cyc}\sqrt{\frac{a^2+pbc}{b^2+c^2}}\,$ where $0\leq p\leq 2$ is: Probably true! Provably true?
Let $p$ be a positive parameter in the range from $0$ to $2$.
Can one prove that
$$2 +\sqrt{\frac p2} \;\leqslant\;\sqrt{\frac{a^2 + pbc}{b^2+c^2}}
\,+\,\sqrt{\frac{b^2 +pca}{c^2+a^2}}\,+\,\sqrt{\...
- 6,517
6
votes
3
answers
206
views
Algebraic inequality $\sum \frac{x^3}{(x+y)(x+z)(x+t)}\geq \frac{1}{2}$
The inequality is
$$\frac{x^3}{(x+y)(x+z)(x+t)}+\frac{y^3}{(y+x)(y+z)(y+t)}+\frac{z^3}{(z+x)(z+y)(z+t)}+\frac{t^3}{(t+x)(t+y)(t+z)}\geq \frac{1}{2},$$
for $x,y,z,t>0$.
It originates from a 3-D ...
- 1,220
6
votes
5
answers
330
views
Bounds on $S = \frac{1}{1001} + \frac{1}{1002}+ \frac{1}{1003}+\dots+\frac{1}{3001}$
$S = \frac{1}{1001} + \frac{1}{1002}+ \frac{1}{1003}+ \dots+\frac{1}{3001}$.
Prove that $\dfrac{29}{27}<S<\dfrac{7}{6}$.
My Attempt:
$S<\dfrac{500}{1000} + \dfrac{500}{1500}+ \dfrac{...
user822140
4
votes
1
answer
98
views
Prove the inequality $\sum_{cyc}\frac{a^3}{b\sqrt{a^3+8}}\ge 1$
Let $a,b,c>0$ and such $a+b+c=3$,show that
$$\sum_{cyc}\dfrac{a^3}{b\sqrt{a^3+8}}\ge 1\tag{1}$$
I tried using Holder's inequality to solve it:
$$\sum_{cyc}\dfrac{a^3}{b\sqrt{a^3+8}}\sum b\sum \...
- 93.6k
4
votes
2
answers
124
views
If $a, b, c, d\in\mathbb R^+, $ then prove that $\displaystyle\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\ge 0.$
My approach: We have:
$\displaystyle\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}$
$\displaystyle=\frac{a+c}{b+c}-1+\frac{b+d}{c+d}-1+\frac{c+a}{d+a}-1+\frac{d+b}{a+b}-1$
$\...
5
votes
4
answers
110
views
Prove that for every $n \in \mathbb{N}$ $\sum\limits_{k=2}^{n}{\frac{1}{k^2}}<1$ [duplicate]
$$\sum\limits_{k=2}^{n}{\frac{1}{k^2}}<1$$
First step would be proving that the statement is true for n=2
On the LHS for $n=2$ we would have $\frac{1}{4}$ therefore the statement is true for $n=...
- 191
5
votes
2
answers
364
views
How to find range $a_{75}$ of the term of the series $a_n=a_{n-1}+ {1 \over {a_{n-1}}} $ [duplicate]
If $a_1=1$ and for n>1$$a_n=a_{n-1}+ {1 \over {a_{n-1}}} $$
$a_{75}$ lies between
(a) (12,15)
(b) (11,12)
(c) (15,18)
Now , in this question, I rewrote, $a_n-a_{n-1} = {1 \over {a_{n-1}}}$, to ...
- 1,766