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Troublesome median in a difficult inequality
JG666   2
N an hour ago by navid
Source: 2022 Spring NSMO Day 2 Problem 3
Determine the minimum value of $\lambda\in\mathbb{R}$, such that for any positive integer $n$ and non-negative reals $x_1, x_2, \cdots, x_n$, the following inequality always holds:
$$\sum_{i=1}^n(m_i-a_i)^2\leqslant \lambda\cdot\sum_{i=1}^nx_i^2,$$Here $m_i$ and $a_i$ denote the median and arithmetic mean of $x_1, x_2, \cdots, x_i$, respectively.

Duanyang ZHANG, High School Affiliated to Renmin University of China
2 replies
JG666
May 22, 2022
navid
an hour ago
Scary Binomial Coefficient Sum
EpicBird08   44
N 2 hours ago by ray66
Source: USAMO 2025/5
Determine, with proof, all positive integers $k$ such that $$\frac{1}{n+1} \sum_{i=0}^n \binom{n}{i}^k$$is an integer for every positive integer $n.$
44 replies
EpicBird08
Mar 21, 2025
ray66
2 hours ago
[$10K+ IN PRIZES] Poolesville Math Tournament (PVMT) 2025
qwerty123456asdfgzxcvb   18
N 2 hours ago by Alex-131
Hi everyone!

After the resounding success of the first three years of PVMT, the Poolesville High School Math Team is excited to announce the fourth annual Poolesville High School Math Tournament (PVMT)! The PVMT team includes a MOPper and multiple USA(J)MO and AIME qualifiers!

PVMT is open to all 6th-9th graders in the country (including rising 10th graders). Students will compete in teams of up to 4 people, and each participant will take three subject tests as well as the team round. The contest is completely free, and will be held virtually on June 7, 2025, from 10:00 AM to 4:00 PM (EST).

Additionally, thanks to our sponsors, we will be awarding approximately $10K+ worth of prizes (including gift cards, Citadel merch, AoPS coupons, Wolfram licenses) to top teams and individuals. More details regarding the actual prizes will be released as we get closer to the competition date.

Further, newly for this year we might run some interesting mini-events, which we will announce closer to the competition date, such as potentially a puzzle hunt and integration bee!

If you would like to register for the competition, the registration form can be found at https://pvmt.org/register.html or https://tinyurl.com/PVMT25.

Additionally, more information about PVMT can be found at https://pvmt.org

If you have any questions not answered in the below FAQ, feel free to ask in this thread or email us at falconsdomath@gmail.com!

We look forward to your participation!

FAQ
18 replies
qwerty123456asdfgzxcvb
Apr 5, 2025
Alex-131
2 hours ago
Introducing a math summer program for middle school students
harry133   2
N 2 hours ago by lovematch13
Introducing IITSP, an online math summer program designed for middle school students over summer.

The program is designed by Professor Shubhrangshu Dasgupta from the Department of Physics at the Indian Institute of Technology Ropar (IIT Ropar).

Please check out the webpage if you are interested in.

https://www.imc-impea.org/IMC/bbs/content.php?co_id=iitsp
2 replies
harry133
Yesterday at 12:52 AM
lovematch13
2 hours ago
Alcumus vs books
UnbeatableJJ   20
N 4 hours ago by SirAppel
If I am aiming for AIME, then JMO afterwards, is Alcumus adequate, or I still need to do the problems on AoPS books?

I got AMC 23 this year, and never took amc 10 before. If I master the alcumus of intermediate algebra (making all of the bars blue). How likely I can qualify for AIME 2026?
20 replies
UnbeatableJJ
Apr 23, 2025
SirAppel
4 hours ago
LTE or Binomial Theorem
P_Groudon   111
N 4 hours ago by heheman
Source: 2020 AIME I #12
Let $n$ be the least positive integer for which $149^n - 2^n$ is divisible by $3^3 \cdot 5^5 \cdot 7^7$. Find the number of positive divisors of $n$.
111 replies
P_Groudon
Mar 12, 2020
heheman
4 hours ago
Mustang Math Recruitment is Open!
MustangMathTournament   4
N 5 hours ago by heheman
The Interest Form for joining Mustang Math is open!

Hello all!

We're Mustang Math, and we are currently recruiting for the 2025-2026 year! If you are a high school or college student and are passionate about promoting an interest in competition math to younger students, you should strongly consider filling out the following form: https://link.mustangmath.com/join. Every member in MM truly has the potential to make a huge impact, no matter your experience!

About Mustang Math

Mustang Math is a nonprofit organization of high school and college volunteers that is dedicated to providing middle schoolers access to challenging, interesting, fun, and collaborative math competitions and resources. Having reached over 4000 U.S. competitors and 1150 international competitors in our first six years, we are excited to expand our team to offer our events to even more mathematically inclined students.

PROJECTS
We have worked on various math-related projects. Our annual team math competition, Mustang Math Tournament (MMT) recently ran. We hosted 8 in-person competitions based in Washington, NorCal, SoCal, Illinois, Georgia, Massachusetts, Nevada and New Jersey, as well as an online competition run nationally. In total, we had almost 900 competitors, and the students had glowing reviews of the event. MMT International will once again be running later in August, and with it, we anticipate our contest to reach over a thousand students.

In our classes, we teach students math in fun and engaging math lessons and help them discover the beauty of mathematics. Our aspiring tech team is working on a variety of unique projects like our website and custom test platform. We also have a newsletter, which, combined with our social media presence, helps to keep the mathematics community engaged with cool puzzles, tidbits, and information about the math world! Our design team ensures all our merch and material is aesthetically pleasing.

Some highlights of this past year include 1000+ students in our classes, AMC10 mock with 150+ participants, our monthly newsletter to a subscriber base of 6000+, creating 8 designs for 800 pieces of physical merchandise, as well as improving our custom website (mustangmath.com, 20k visits) and test-taking platform (comp.mt, 6500+ users).

Why Join Mustang Math?

As a non-profit organization on the rise, there are numerous opportunities for volunteers to share ideas and suggest projects that they are interested in. Through our organizational structure, members who are committed have the opportunity to become a part of the leadership team. Overall, working in the Mustang Math team is both a fun and fulfilling experience where volunteers are able to pursue their passion all while learning how to take initiative and work with peers. We welcome everyone interested in joining!

More Information

To learn more, visit https://link.mustangmath.com/RecruitmentInfo. If you have any questions or concerns, please email us at contact@mustangmath.com.

https://link.mustangmath.com/join
4 replies
MustangMathTournament
May 24, 2025
heheman
5 hours ago
Wrong Answer on a Street Math Challenge
miguel00   17
N Yesterday at 4:09 PM by miguel00
Hello AoPS Community,

I was just watching this video link (those of you that are Korean, you should watch it!) but I came across a pretty hard vector geometry problem (keep in mind contestants have to solve this problem in 5 minutes). No one got this problem (no surprise there) but I am posting because I actually think the answer he gave is wrong.

So the problem goes like this: Referencing the diagram attached, there are three externally tangent unit circles $C_1, C_2, C_3$ on a plane with centers $O_1, O_2, O_3$, respectively. $H$ is feet of the perpendicular from $O_1$ to $O_2O_3$ and $A$ and $B$ are intersections of line $O_1H$ with circle $C_1$. Points $P$ and $Q$ can move around the circle $C_2$ and $C_3$, respectively. Find the maximum possible value of $|\overrightarrow{AQ}+\overrightarrow{PB}|$.


I got my answer but the video said their 1st answer but they later corrected it on the comments to their 2nd answer. I'll let you guys attempt the problem and will give my solution shortly after. Thanks in advance!

-miguel00
17 replies
miguel00
Yesterday at 1:37 AM
miguel00
Yesterday at 4:09 PM
2 headed arrows usage
mathprodigy2011   1
N Tuesday at 11:30 PM by alcumusftwgrind
Source: 2003 USAMO 4
I can't upload the file but I was working with someone on 2003 USAMO p4. When we saw "if and only if" I thought it meant we have to prove it both directions. However, when we looked at Evan Chen's solution after writing it out; Evan Chen used double headed arrows and left it at that. My question is, how did he use them and how do I know when I can or can not use them?
1 reply
mathprodigy2011
Tuesday at 11:09 PM
alcumusftwgrind
Tuesday at 11:30 PM
4th grader qual JMO
HCM2001   42
N Tuesday at 6:58 PM by BS2012
i mean.. whattttt??? just found out about this.. is he on aops? (i'm sure he is) where are you orz lol..
https://www.mathschool.com/blog/results/celebrating-success-douglas-zhang-is-rsm-s-youngest-usajmo-qualifier
42 replies
HCM2001
May 22, 2025
BS2012
Tuesday at 6:58 PM
Do you need to attend mop
averageguy   5
N Tuesday at 5:55 PM by babyzombievillager
So I got accepted into a summer program and already paid the fee of around $5000 dollars. It's for 8 weeks (my entire summer) and it's in person. I have a few questions
1. If I was to make MOP this year am I forced to attend?
2.If I don't attend the program but still qualify can I still put on my college application that I qualified for MOP or can you only put MOP qualifier if you actually attend the program.
5 replies
averageguy
Mar 5, 2025
babyzombievillager
Tuesday at 5:55 PM
Equality occurs in strange points
arqady   10
N Mar 29, 2025 by sqing
Let $a$, $b$ and $c$ are non-negatives such that $a+b+c\neq0$. Prove that:
1) \[\sum_{cyc}\sqrt{a^2+b^2}\leq\frac{11(a^2+b^2+c^2)+7(ab+ac+bc)}{3\sqrt2(a+b+c)}\]
2)\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]
3)\[\sum_{cyc}\sqrt{4a^2+ab+4b^2}\leq\frac{5(a^2+b^2+c^2)+4(ab+ac+bc)}{a+b+c}\]
10 replies
arqady
Aug 28, 2011
sqing
Mar 29, 2025
Equality occurs in strange points
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arqady
30260 posts
#1 • 3 Y
Y by luofangxiang, Adventure10, Mango247
Let $a$, $b$ and $c$ are non-negatives such that $a+b+c\neq0$. Prove that:
1) \[\sum_{cyc}\sqrt{a^2+b^2}\leq\frac{11(a^2+b^2+c^2)+7(ab+ac+bc)}{3\sqrt2(a+b+c)}\]
2)\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]
3)\[\sum_{cyc}\sqrt{4a^2+ab+4b^2}\leq\frac{5(a^2+b^2+c^2)+4(ab+ac+bc)}{a+b+c}\]
This post has been edited 1 time. Last edited by arqady, Aug 29, 2011, 1:03 PM
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Tourish
663 posts
#2 • 2 Y
Y by Adventure10, Mango247
I think they are good practice for Cauchy inequality and pqr method :lol:
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arqady
30260 posts
#3 • 2 Y
Y by Adventure10, Mango247
Tourish wrote:
I think they are good practice for Cauchy inequality and pqr method :lol:
What is your proof for the second inequality? I think, it's not so trivially. :wink:
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Tourish
663 posts
#4 • 1 Y
Y by Adventure10
For the second one it is very trival if the constant is $9$(it is just your typo?), and in fact I think we have
\[\sum_{cyc}{\sqrt{a^2+7ab+b^2}}\leq\frac{2(a^2+b^2+c^2)+7(ab+bc+ca)}{a+b+c}\]
But I haven't proved it yet :blush:
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arqady
30260 posts
#5 • 2 Y
Y by Adventure10, Mango247
Thank you, Tourish! I have fixed it.
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Tourish
663 posts
#6 • 2 Y
Y by Adventure10 and 1 other user
My proof for the second one: if there are two variables of $a,b,c$ are equal to $0$, the original inequality is obvious. Now consider $ab+bc+ca>0$. The inequality is equavilent to
\[\sum{\left(\frac{3}{2}(a+b)-\sqrt{a^2+7ab+b^2}\right)}\geq 3\sum{a}-\frac{2(a^2+b^2+c^2)+7(ab+bc+ca)}{a+b+c}\]
\[\Longleftrightarrow \sum{\frac{5(a-b)^2}{6(a+b)+4\sqrt{a^2+7ab+b^2}}}\geq \frac{\sum{(a-b)^2}}{2(a+b+c)}\]
which can be writen as $S_a(b-c)^2+S_b(c-a)^2+S_c(a-b)^2\geq 0$, where
\[S_c=\frac{5}{6(a+b)+4\sqrt{a^2+7ab+b^2}}-\frac{1}{2(a+b+c)}\]
and $S_a, S_b$ can be expressed similarly. Assume that $a\geq b\geq c$. First, let's prove that $S_b\geq 0$, that is
\[\frac{5}{6(c+a)+4\sqrt{c^2+7ca+a^2}}\geq \frac{1}{2(a+b+c)}\]
\[\Longleftrightarrow 2a+2c+5b\geq 2\sqrt{c^2+7ca+a^2}\]
\[\Longleftrightarrow -20ac+20ab+20bc+25b^2\geq 0\]
which is true since $a\geq b\geq c$. We can prove $S_a\geq 0$ as the same way. Now it suffices to show that $S_b+S_c\geq 0$, which is
\[\frac{5}{6(a+b)+4\sqrt{a^2+7ab+b^2}}+\frac{5}{6(c+a)+4\sqrt{c^2+7ca+a^2}}\geq \frac{1}{a+b+c}\]
By Cauchy-Schwarz inequality we only need to prove
\[\frac{20}{12a+6b+6c+4\sqrt{a^2+7ab+b^2}+4\sqrt{c^2+7ca+a^2}}\geq \frac{1}{a+b+c}\]
\[\Longleftrightarrow 4a+7b+7c\geq 2(\sqrt{a^2+7ab+b^2}+\sqrt{c^2+7ca+a^2})\]
By Cauchy-Schwarz again, we need to prove that
\[(4a+7b+7c)^2\geq 8(2a^2+b^2+c^2+7ab+7ac)\]
which is obvious true after expanding. Then by SOS theory, our inequality is true. Equal holds only when $a=b=c$ or $a=b=0$ and its permutations .
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can_hang2007
2948 posts
#7 • 5 Y
Y by Adventure10, Mango247, MS_asdfgzxcvb, and 2 other users
arqady wrote:
Let $a$, $b$ and $c$ are non-negatives such that $a+b+c\neq0$. Prove that:
2)\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]
I will post my solution to this one.

Without loss of generality, assume that $a \ge b \ge c.$ If $b=c=0,$ then the inequality is trivial, so we only need to consider $b>0.$ Now, using the AM-GM inequality, we get
$2\sqrt{a^2+7ab+b^2} \le a+2b+\frac{a^2+7ab+b^2}{a+2b}=2a+2b+\frac{5ab+b^2}{a+2b},$
$2\sqrt{a^2+7ac+c^2} \le a+2c+\frac{a^2+7ac+c^2}{a+2c} =2a+2c+\frac{5ac+c^2}{a+2c}$
and
$2\sqrt{b^2+7bc+c^2} \le b+2c+\frac{b^2+7bc+c^2}{b+2c}=2b+2c+\frac{5bc+c^2}{b+2c}.$
Therefore, it suffices to prove that
$\frac{5ab+b^2}{a+2b}+\frac{5ac+c^2}{a+2c}+\frac{5bc+c^2}{b+2c} \le \frac{4(a^2+b^2+c^2)+14(ab+bc+ca)}{a+b+c}-4(a+b+c),$
or
$\frac{5ab+b^2}{a+2b}+\frac{5ac+c^2}{a+2c}+\frac{5bc+c^2}{b+2c} \le \frac{6(ab+bc+ca)}{a+b+c}.$
Multiplying each side by $2$ with noting that $\frac{2(5ab+b^2)}{a+2b}=b+\frac{9ab}{a+2b},$ we can rewrite the above inequality as
$b+2c+\frac{9ab}{a+2b}+\frac{9ac}{a+2c}+\frac{9bc}{b+2c} \le \frac{12(ab+bc+ca)}{a+b+c},$
or
$(b+2c)(a+b+c)+\frac{9ab(a+b+c)}{a+2b}+\frac{9ac(a+b+c)}{a+2c}+\frac{9bc(a+b+c)}{b+2c}\le $
$\le 12(ab+bc+ca).$
This rewrites as
$9a\left[\frac{b(a+b+c)}{a+2b}+\frac{c(a+b+c)}{a+2c}- b-c\right] +(a+b+c)\left(\frac{9bc}{b+2c} -2b-c\right) \le $
$\le 12(ab+bc+ca)- 9a(b+c)-(2b+c)(a+b+c)-(b+2c)(a+b+c).$
After some simple computations, we get
$-\frac{9a^2(b-c)^2}{(a+2b)(a+2c)} -\frac{2(a+b+c)(b-c)^2}{b+2c} \le -3(b-c)^2, $
or
$(b-c)^2\left[\frac{9a^2}{(a+2b)(a+2c)}+\frac{2(a+b+c)}{b+2c} -3\right] \ge 0,$
which is true because $\frac{9a^2}{(a+2b)(a+2c)} \ge 1$ and $\frac{2(a+b+c)}{b+2c} \ge 2.$
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luofangxiang
4613 posts
#8 • 2 Y
Y by Adventure10, Mango247
arqady wrote:
Let $a$, $b$ and $c$ are non-negatives such that $a+b+c\neq0$. Prove that:
1) \[\sum_{cyc}\sqrt{a^2+b^2}\leq\frac{11(a^2+b^2+c^2)+7(ab+ac+bc)}{3\sqrt2(a+b+c)}\]2)\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]3)\[\sum_{cyc}\sqrt{4a^2+ab+4b^2}\leq\frac{5(a^2+b^2+c^2)+4(ab+ac+bc)}{a+b+c}\]

//cdn.artofproblemsolving.com/images/8/a/b/8ab4482f5069c0a532962fed49e9a543a5313ea3.jpg
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luofangxiang
4613 posts
#9 • 2 Y
Y by Adventure10, Mango247
(1)//cdn.artofproblemsolving.com/images/8/9/e/89ec700132c4078994bb73bf252d9f2ecdc7f98d.jpg
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manutd03
350 posts
#10 • 1 Y
Y by Adventure10
Can we solve the first one by S.O.S theorem, i think i got some trouble with Sa,Sb,Sc
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sqing
42484 posts
#11
Y by
can_hang2007 wrote:
*
$\frac{5ab+b^2}{a+2b}+\frac{5ac+c^2}{a+2c}+\frac{5bc+c^2}{b+2c} \le \frac{6(ab+bc+ca)}{a+b+c}.$
**
Let $ a,b,c\ge 0 $ and $  ab+bc+ca>0 .$ Prove that
$$\frac{a^2+5bc}{b+c}+\frac{b^2+5ca}{c+a} +\frac{c^2+5ab}{a+b}\le \frac{9(ab+bc+ca)}{a+b+c} $$
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