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k a June Highlights and 2025 AoPS Online Class Information
jlacosta   0
Jun 2, 2025
Congratulations to all the mathletes who competed at National MATHCOUNTS! If you missed the exciting Countdown Round, you can watch the video at this link. Are you interested in training for MATHCOUNTS or AMC 10 contests? How would you like to train for these math competitions in half the time? We have accelerated sections which meet twice per week instead of once starting on July 8th (7:30pm ET). These sections fill quickly so enroll today!

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0 replies
jlacosta
Jun 2, 2025
0 replies
two subsets with no fewer than four common elements.
micliva   41
N 3 minutes ago by happypi31415
Source: All-Russian Olympiad 1996, Grade 9, First Day, Problem 4
In the Duma there are 1600 delegates, who have formed 16000 committees of 80 persons each. Prove that one can find two committees having no fewer than four common members.

A. Skopenkov
41 replies
micliva
Apr 18, 2013
happypi31415
3 minutes ago
King's Constrained Walk
Hellowings   4
N 10 minutes ago by Diamond-jumper76
Source: Own
Given an n x n chessboard, with a king starting at any square, the king's task is to visit each square in the board exactly once (essentially an open path); this king moves how a king in chess would.
However, we are allowed to place k numbers on the board of any value such that for each number A we placed on the board, the king must be in the position of that number A on its Ath square in its journey, with the starting square as its 1st square.
Suppose after we placed k numbers, there is one and only one way to complete the king's task (this includes placing the king in a starting square), find the minimum value of k set by n.

Should've put one of its tag as "Open problem"; I have no idea how to tackle this problem either.
4 replies
Hellowings
May 30, 2025
Diamond-jumper76
10 minutes ago
polygon's area doesn't add much when combined with its centric symmetry
mathematics2003   8
N 29 minutes ago by CrazyInMath
Source: 2021ChinaTST test4 day2 P2
Find the smallest real $\alpha$, such that for any convex polygon $P$ with area $1$, there exist a point $M$ in the plane, such that the area of convex hull of $P\cup Q$ is at most $\alpha$, where $Q$ denotes the image of $P$ under central symmetry with respect to $M$.
8 replies
mathematics2003
Apr 14, 2021
CrazyInMath
29 minutes ago
Inequality
doquocchinh   0
an hour ago
Let \( a, b, c \) be non-negative real numbers satisfying \( ab + bc + ca > 0 \) and \( a + b + c = 3 \). Prove that:
\[
3(ab + bc + ca) \left( \frac{a + b}{b + c} + \frac{b + c}{c + a} + \frac{c + a}{a + b} \right) \geq 23 + \frac{t^3(21 - 5t)}{2(3 - t)}
\]where \( t = \sqrt[3]{abc} \).
0 replies
doquocchinh
an hour ago
0 replies
No more topics!
hard problem
Cobedangiu   15
N May 1, 2025 by arqady
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$
15 replies
Cobedangiu
Apr 21, 2025
arqady
May 1, 2025
hard problem
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Cobedangiu
70 posts
#1 • 1 Y
Y by RainbowJessa
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$
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m4thbl3nd3r
294 posts
#2 • 1 Y
Y by RainbowJessa
Cobedangiu wrote:
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$

Tangent line :whistling:
Z K Y
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giangtruong13
156 posts
#3 • 1 Y
Y by RainbowJessa
giangtruong13 wrote:
Bài này giống với bài BĐT trong đề thi HSG Thái Bình năm 2024-2025
Z K Y
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Jackson0423
114 posts
#4 • 1 Y
Y by RainbowJessa
use the constants
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Cobedangiu
70 posts
#5 • 1 Y
Y by RainbowJessa
giangtruong13 wrote:
giangtruong13 wrote:
Bài này giống với bài BĐT trong đề thi HSG Thái Bình năm 2024-2025
nó có thể giải đc chỉ với Schwarz .-.
Z K Y
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arqady
30263 posts
#6 • 1 Y
Y by RainbowJessa
Cobedangiu wrote:
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$
It's just Popoviciu for $f(x)=\frac{1}{x}$ on $(0,+\infty).$
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IceyCold
212 posts
#7 • 1 Y
Y by RainbowJessa
m4thbl3nd3r wrote:
Cobedangiu wrote:
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$

Tangent line :whistling:

mhmm,Tangent Line
I like
Z K Y
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arqady
30263 posts
#8 • 1 Y
Y by RainbowJessa
IceyCold wrote:
mhmm,Tangent Line
I like
Did you try? I think, it does not help here.
Z K Y
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ReticulatedPython
719 posts
#9 • 1 Y
Y by RainbowJessa
Interesting problem. I suspect that AM-GM might be applicable here, since equality is achieved at $a=b=c=1$ (which is the AM-GM equality condition).
This post has been edited 1 time. Last edited by ReticulatedPython, Apr 24, 2025, 3:04 PM
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IceyCold
212 posts
#10 • 1 Y
Y by RainbowJessa
arqady wrote:
IceyCold wrote:
mhmm,Tangent Line
I like
Did you try? I think, it does not help here.

It was one of our test.The graders marked my method correct,so I hope darn well I am right lol-
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Cobedangiu
70 posts
#11 • 1 Y
Y by RainbowJessa
IceyCold wrote:
arqady wrote:
IceyCold wrote:
mhmm,Tangent Line
I like
Did you try? I think, it does not help here.

It was one of our test.The graders marked my method correct,so I hope darn well I am right lol-

Can you write your method?
Z K Y
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ReticulatedPython
719 posts
#12 • 1 Y
Y by RainbowJessa
IceyCold wrote:
arqady wrote:
IceyCold wrote:
mhmm,Tangent Line
I like
Did you try? I think, it does not help here.

It was one of our test.The graders marked my method correct,so I hope darn well I am right lol-

Yeah can you share the method with us?
Z K Y
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Edward_Tur
132 posts
#13
Y by
Cobedangiu wrote:
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$

$a=\frac{3x}{x+y+z},...$
$\sum_{sym} x^4y^2-x^4yz+x^3y^3-x^2y^2z^2\ge0.$
This post has been edited 1 time. Last edited by Edward_Tur, Apr 28, 2025, 7:44 PM
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IceyCold
212 posts
#14
Y by
Cobedangiu wrote:
Can you write your method?
Fakesolve
This post has been edited 3 times. Last edited by IceyCold, Apr 30, 2025, 1:36 AM
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IceyCold
212 posts
#15
Y by
IceyCold wrote:
Cobedangiu wrote:
Can you write your method?
Show that $\frac{4}{3-c} \le \frac{1}{c} -2c + 3 $.

This is equivalent to $\frac{(c-1)^2(2c+3)}{c(c-3)} \le 0$,trivially true.

Never mind,I see the issue now.
Sorry for a waste of time-
This post has been edited 1 time. Last edited by IceyCold, Apr 30, 2025, 1:34 AM
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arqady
30263 posts
#17
Y by
ReticulatedPython wrote:

Yeah can you share the method with us?
See here
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