Art of Problem Solving
AoPS Online AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy AoPS Academy
Small live classes for advanced math
and language arts learners in grades 1-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Join our free webinar April 22 to learn about competitive programming!
inequalities
B
No tags match your search
M
G
Topic
First Poster
Last Poster
H
Stronger inequality than an old result
KhuongTrang   20
N 5 minutes ago by KhuongTrang
Source: own, inspired
Problem. Find the best constant $k$ satisfying $$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$holds for all $a,b,c\ge 0: ab+bc+ca>0.$
20 replies
KhuongTrang
Aug 1, 2024
KhuongTrang
5 minutes ago
J
H
Two very hard parallel
jayme   3
N 35 minutes ago by jayme
Source: own inspired by EGMO
Dear Mathlinkers,

1. ABC a triangle
2. D, E two point on the segment BC so that BD = DE= EC
3. M, N the midpoint of ED, AE
4. H the orthocenter of the acutangle triangle ADE
5. 1, 2 the circumcircle of the triangle DHM, EHN
6. P, Q the second point of intersection of 1 and BM, 2 and CN
7. U, V the second points of intersection of 2 and MN, PQ.

Prove : UV is parallel to PM.

Sincerely
Jean-Louis
3 replies
jayme
Yesterday at 12:46 PM
jayme
35 minutes ago
J
H
Inequality with n-gon sides
mihaig   3
N 36 minutes ago by mihaig
Source: VL
If $a_1,a_2,\ldots, a_n~(n\geq3)$ are are the lengths of the sides of a $n-$gon such that
$$\sum_{i=1}^{n}{a_i}=1,$$then
$$(n-2)\left[\sum_{i=1}^{n}{\frac{a_i^2}{(1-a_i)^2}}-\frac n{(n-1)^2}\right]\geq(2n-1)\left(\sum_{i=1}^{n}{\frac{a_i}{1-a_i}}-\frac n{n-1}\right)^2.$$
When do we have equality?

(V. Cîrtoaje and L. Giugiuc, 2021)
3 replies
mihaig
Feb 25, 2022
mihaig
36 minutes ago
J
H
Advanced topics in Inequalities
va2010   23
N 37 minutes ago by Novmath
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
23 replies
1 viewing
va2010
Mar 7, 2015
Novmath
37 minutes ago
J
H
JBMO TST Bosnia and Herzegovina 2022 P3
Motion   7
N 43 minutes ago by cafer2861
Source: JBMO TST Bosnia and Herzegovina 2022
Let $ABC$ be an acute triangle. Tangents on the circumscribed circle of triangle $ABC$ at points $B$ and $C$ intersect at point $T$. Let $D$ and $E$ be a foot of the altitudes from $T$ onto $AB$ and $AC$ and let $M$ be the midpoint of $BC$. Prove:
A) Prove that $M$ is the orthocenter of the triangle $ADE$.
B) Prove that $TM$ cuts $DE$ in half.
7 replies
Motion
May 21, 2022
cafer2861
43 minutes ago
J
H
hard problem
Cobedangiu   5
N an hour ago by arqady
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$
5 replies
Cobedangiu
Yesterday at 1:51 PM
arqady
an hour ago
J
H
density over modulo M
SomeGuy3335   3
N 2 hours ago by ja.
Let $M$ be a positive integer and let $\alpha$ be an irrational number. Show that for every integer $0\leq a < M$, there exists a positive integer $n$ such that $M \mid \lfloor{n \alpha}\rfloor-a$.
3 replies
SomeGuy3335
Apr 20, 2025
ja.
2 hours ago
J
H
Diophantine equation !
ComplexPhi   5
N 2 hours ago by aops.c.c.
Source: Romania JBMO TST 2015 Day 1 Problem 4
Solve in nonnegative integers the following equation :
$$21^x+4^y=z^2$$
5 replies
ComplexPhi
May 14, 2015
aops.c.c.
2 hours ago
J
H
Combo problem
soryn   0
3 hours ago
The school A has m1 boys and m2 girls, and ,the school B has n1 boys and n2 girls. Each school is represented by one team formed by p students,boys and girls. If f(k) is the number of cases for which,the twice schools has,togheter k girls, fund f(k) and the valute of k, for which f(k) is maximum.
0 replies
soryn
3 hours ago
0 replies
J
H
Parity and sets
betongblander   7
N 3 hours ago by ihategeo_1969
Source: Brazil National Olympiad 2020 5 Level 3
Let $n$ and $k$ be positive integers with $k$ $\le$ $n$. In a group of $n$ people, each one or always
speak the truth or always lie. Arnaldo can ask questions for any of these people
provided these questions are of the type: “In set $A$, what is the parity of people who speak to
true? ”, where $A$ is a subset of size $ k$ of the set of $n$ people. The answer can only
be “$even$” or “$odd$”.
a) For which values of $n$ and $k$ is it possible to determine which people speak the truth and
which people always lie?
b) What is the minimum number of questions required to determine which people
speak the truth and which people always lie, when that number is finite?
7 replies
betongblander
Mar 18, 2021
ihategeo_1969
3 hours ago
J
H
Mount Inequality erupts on a sequence :o
GrantStar   88
N 3 hours ago by Nari_Tom
Source: 2023 IMO P4
Let $x_1,x_2,\dots,x_{2023}$ be pairwise different positive real numbers such that
\[a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}\]is an integer for every $n=1,2,\dots,2023.$ Prove that $a_{2023} \geqslant 3034.$
88 replies
GrantStar
Jul 9, 2023
Nari_Tom
3 hours ago
J
H
J
H
China Western Mathematical Olympiad 2015 ,Problem 3
sqing   8
N Feb 5, 2025 by shanelin-sigma
Source: China Yinchuan Aug 16, 2015
Let the integer $n \ge 2$ , and $x_1,x_2,\cdots,x_n $ be positive real numbers such that $\sum_{i=1}^nx_i=1$ .Prove that$$\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{1\le i<j\le n} x_ix_j\right)\le \frac{n}{2}.$$
8 replies
sqing
Aug 16, 2015
shanelin-sigma
Feb 5, 2025
J
China Western Mathematical Olympiad 2015 ,Problem 3
G H J
Reveal topic tags
inequalities
L
G H BBookmark kLocked kLocked NReply
Source: China Yinchuan Aug 16, 2015
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
41773 posts
sqing
#1 h  Aug 16, 2015, 10:33 AM • 1 Y
Y by Adventure10
Let the integer $n \ge 2$ , and $x_1,x_2,\cdots,x_n $ be positive real numbers such that $\sum_{i=1}^nx_i=1$ .Prove that$$\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{1\le i<j\le n} x_ix_j\right)\le \frac{n}{2}.$$
This post has been edited 1 time. Last edited by sqing, Aug 16, 2015, 11:25 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MariusStanean
655 posts
MariusStanean
#2 h  Aug 16, 2015, 5:41 PM • 4 Y
Y by rkm0959, zsgvivo, Adventure10, Mango247
Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have
$$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
41773 posts
sqing
#3 h  Aug 16, 2015, 10:25 PM • 1 Y
Y by Adventure10
Very nice.
Thanks.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
zsgvivo
28 posts
zsgvivo
#4 h  Jul 24, 2019, 1:03 AM • 2 Y
Y by Adventure10, Mango247
MariusStanean wrote:
Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have
$$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$

what is Cebasev inequality
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
41773 posts
sqing
#5 h  Jul 24, 2019, 1:24 AM • 2 Y
Y by zsgvivo, Adventure10
zsgvivo wrote:
MariusStanean wrote:
Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have
$$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$

what is Cebasev inequality
https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
zsgvivo
28 posts
zsgvivo
#6 h  Jul 24, 2019, 1:40 AM • 1 Y
Y by Adventure10
sqing wrote:
zsgvivo wrote:
MariusStanean wrote:
Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have
$$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$

what is Cebasev inequality
https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality

thanks!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bel.jad5
3750 posts
bel.jad5
#7 h  Jul 24, 2019, 11:17 AM • 2 Y
Y by Adventure10, Mango247
@fair enogh...thanks
This post has been edited 1 time. Last edited by bel.jad5, Jul 24, 2019, 1:03 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
WolfusA
1900 posts
WolfusA
#8 h  Jul 24, 2019, 11:57 AM • 2 Y
Y by Adventure10, Mango247
For $a+b<1, a,b\ge 0$ holds $\frac{1}{1-a}\le\frac{1}{1-b}\iff a\le b$ and $a(1-a)\le b(1-b)\iff a-b\le (a-b)(a+b)\iff (1-a-b)(a-b)\le0\iff a\le b$ which proves sequences $\left(\frac{1}{1-x_i}\right)$, $\left(x_i(1-x_i)\right)$ are non-decreasing for $0<x_1\le x_2\le\ldots \le x_n, x_1+x_2+...+x_n=1$ so we can use Chebyshev
This post has been edited 1 time. Last edited by WolfusA, Jul 24, 2019, 11:57 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shanelin-sigma
155 posts
shanelin-sigma
#12 h  Feb 5, 2025, 12:27 AM
Y by
MariusStanean wrote:
Suppose that $x_1\le x_2\le\ldots \le x_n$ and with Cebasev inequality we have
$$2\cdot LHS=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(1-\sum_{i=1}^n x_i^2\right)=\left(\sum_{i=1}^n\frac{1}{1-x_i}\right)\left(\sum_{i=1}^n x_i(1-x_i)\right)\le n\sum_{i=1}^n x_i=n$$

But $x_i(1-x_i)$ isn’t always increase as $x_i$ increase
Should we check the case if $x_n> \frac 12$?
This post has been edited 1 time. Last edited by shanelin-sigma, Feb 5, 2025, 12:29 AM
Z K Y
N Quick Reply
G
H
=
a