All Questions
46
questions
15
votes
2
answers
608
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
$$2 +\sqrt{\frac p2} \;\leqslant\;\sqrt{\frac{a^2 + pbc}{b^2+c^2}}
\,+\,\sqrt{\frac{b^2 +pca}{c^2+a^2}}\,+\,\sqrt{\...
- 6,528
12
votes
2
answers
937
views
$\sum\limits_{i=1}^n \frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$, for all $x_i>0.$
Can you prove the following new inequality? I found it experimentally.
Prove that, for all $x_1,x_2,\ldots,x_n>0$, it holds that
$$\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod\limits _{j=...
- 2,040
10
votes
3
answers
718
views
Prove that $\sum\limits_{k=1}^n \frac{1}{k^2+3k+1}$ is bounded above by $\frac{13}{20}$
I want ask a question about a sum. The exercise is as follows:
Prove the following inequality for every $n \geq 1$:
$$\sum\limits_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{13}{20} .$$
- 101
7
votes
1
answer
244
views
Prove that $\frac{1}{1+a_1+a_1a_2}+\frac{1}{1+a_2+a_2a_3}+\cdots+\frac{1}{1+a_{n-1}+a_{n-1}a_n}+\frac{1}{1+a_n+a_na_1}>1.$
If $n > 3$ and $a_1,a_2,\ldots,a_n$ are positive real numbers with $a_1a_2\cdots a_n = 1$, prove that $$\dfrac{1}{1+a_1+a_1a_2}+\dfrac{1}{1+a_2+a_2a_3}+\cdots+\dfrac{1}{1+a_{n-1}+a_{n-1}a_n}+\dfrac{...
- 15.6k
7
votes
1
answer
184
views
Show that $|x_{k+1}-x_k| \leq 1$ (for $0<k<n$) implies $\sum_{k=1}^n |x_k| - \left|\sum_{k=1}^n x_k\right|\leq\lceil(n^2-1)/4\rceil$.
Let $n\ge 1$ be a positive integer and let $x_1,\cdots, x_n$ be real numbers so that $|x_{k+1}-x_k|\leq 1$ for $k=1,2,\cdots, n-1$. Show $$\sum_{k=1}^n |x_k| - \left|\sum_{k=1}^n x_k\right|\leq \left\...
- 1,837
6
votes
3
answers
208
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 ...
- 1,250
6
votes
5
answers
333
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
6
votes
2
answers
144
views
How to show $\sum\limits_{r=0}^n \frac{1}{r!} \lt\left (1 + \frac{1}{n}\right)^{n+1}$ for all $n \ge 1$?
Using the binomial expansion, it is quite is easy to show that $$\left(1+\frac{1}{n}\right)^n \le \sum_{r=0}^{n} \frac{1}{r!} $$ for all $n\in\mathbb{Z^+}$, with equality holds when $n=1.$ (Can it be ...
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
5
votes
3
answers
450
views
Non-induction proof of $2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1$
Prove that $$2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1.$$
After playing around with the sum, I couldn't get anywhere so I proved inequalities by induction. I'm however ...
- 2,916
5
votes
2
answers
520
views
Prove that $\sum_{i=1}^n\frac{x_i}{\sqrt[nr]{x_i^{nr}+(n^{nr}-1)\prod_{j=1}^nx^r_j}} \ge 1$ for all $x_i>0$ and $r \geq \frac{1}{n}$.
Prove that, for all $x_1,x_2,\ldots,x_n>0$ and $r \geq \frac{1}{n}$, it holds that
$$\sum_{i=1}^n\frac{x_i}{\sqrt[nr]{x_i^{nr}+(n^{nr}-1)\prod \limits_{j=1}^nx^r_j}} \ge 1.$$
This is a slightly ...
- 2,040
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:
$\displaystyle\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}$
$\displaystyle=\frac{a+c}{b+c}-1+\frac{b+d}{c+d}-1+\frac{c+a}{d+a}-1+\frac{d+b}{a+b}-1$
$\...
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:
$$\sum_{cyc}\dfrac{a^3}{b\sqrt{a^3+8}}\sum b\sum \...
- 93.6k
3
votes
2
answers
196
views
Inequality $\frac{1}{a+b}+\frac{1}{a+2b}+...+\frac{1}{a+nb}<\frac{n}{\sqrt{a\left( a+nb \right)}}$
Let $a,b\in \mathbb{R+}$ and $n\in \mathbb{N}$. Prove that:
$$\frac{1}{a+b}+\frac{1}{a+2b}+...+\frac{1}{a+nb}<\frac{n}{\sqrt{a\left( a+nb \right)}}$$
I have a solution using induction, but ...
- 765