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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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0 replies
jlacosta
Mar 2, 2025
0 replies
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A property of divisors
rightways   8
N 7 minutes ago by de-Kirschbaum
Source: Kazakhstan NMO 2016, P1
Prove that one can arrange all positive divisors of any given positive integer around a circle so that for any two neighboring numbers one is divisible by another.
8 replies
rightways
Mar 17, 2016
de-Kirschbaum
7 minutes ago
J
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Nice problem
hanzo.ei   1
N 33 minutes ago by Mathzeus1024
Find all functions \( f: \mathbb{R} \to \mathbb{R} \) such that
\[ f(xy) = f(x)f(y) \;-\; f(x + y) \;+\; 1, \quad \forall x, y \in \mathbb{R}. \]
1 reply
1 viewing
hanzo.ei
2 hours ago
Mathzeus1024
33 minutes ago
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ortho conf DEF, radius MD, intersect ME,MF, collinear H,K,L
star-1ord   0
an hour ago
Source: Estonia Final Round 2025 12-3
Let $ABC$ be an acute-angled triangle with $|AB|<|AC|$. The altitudes $AD,BE$ and $CF$ intersect at $H$. Let $M$ be the midpoint of $BC$. Point $K$ is chosen on the extension of $EM$ beyond $M$ and point $L$ is chosen on the segment $FM$ such that $|MK|=|ML|=|MD|$. Prove that points $K, L$ and $H$ are collinear.

a little harder version
0 replies
star-1ord
an hour ago
0 replies
J
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Funny system of equations in three variables
Tintarn   10
N 2 hours ago by Marcus_Zhang
Source: Baltic Way 2020, Problem 5
Find all real numbers $x,y,z$ so that
\begin{align*} x^2 y + y^2 z + z^2 &= 0 \\ z^3 + z^2 y + z y^3 + x^2 y &= \frac{1}{4}(x^4 + y^4). \end{align*}
10 replies
Tintarn
Nov 14, 2020
Marcus_Zhang
2 hours ago
J
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A hard inequality
JK1603JK   2
N Today at 2:25 AM by sqing
Let a,b,c\ge 0: a+b+c=3. Prove \frac{1}{abc}+\frac{12}{a^2b+b^2c+c^2a}\ge 5.
2 replies
JK1603JK
Today at 1:40 AM
sqing
Today at 2:25 AM
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Inequalities
sqing   3
N Yesterday at 4:13 PM by sqing
Let $ a,b> 0$ and $ a+b=1 . $ Prove that
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4096}}{1+ ka^7b^7}$$Where $\frac{8192}{3}\geq k>0 .$
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{\frac{14}{3}}{1+ \frac{8192}{3}a^7b^7}$$
3 replies
sqing
Yesterday at 3:14 PM
sqing
Yesterday at 4:13 PM
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Inequalities from SXTX
sqing   11
N Yesterday at 3:47 PM by byron-aj-tom
T702. Let $ a,b,c>0 $ and $ a+2b+3c=\sqrt{13}. $ Prove that $$ \sqrt{a^2+1} +2\sqrt{b^2+1} +3\sqrt{c^2+1} \geq 7$$S
T703. Let $ a,b $ be real numbers such that $ a+b\neq 0. $. Find the minimum of $ a^2+b^2+(\frac{1-ab}{a+b} )^2.$
T704. Let $ a,b,c>0 $ and $ a+b+c=3. $ Prove that $$ \frac{a^2+7}{(c+a)(a+b)} + \frac{b^2+7}{(a+b)(b+c)} +\frac{c^2+7}{(b+c)(c+a)} \geq 6$$S
11 replies
sqing
Feb 18, 2025
byron-aj-tom
Yesterday at 3:47 PM
J
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a^{2000}+b^{2000}=a^{1998}+b^{1998} (Greece Junior 1999 p1)
parmenides51   2
N Yesterday at 12:20 PM by ali123456
Show that if $a,b$ are positive real numbers such that $a^{2000}+b^{2000}=a^{1998}+b^{1998}$ then $a^2+b^2 \le 2$.
2 replies
parmenides51
Mar 17, 2020
ali123456
Yesterday at 12:20 PM
J
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An inequality
jokehim   3
N Yesterday at 10:58 AM by Indpsolver
Let $a,b,c \in \mathbb{R}: a+b+c=3$ then prove $$\color{black}{\frac{a^2}{a^{2}-2a+3}+\frac{b^2}{b^{2}-2b+3}+\frac{c^2}{c^{2}-2c+3}\ge \frac{3}{2}.}$$
3 replies
jokehim
Mar 21, 2025
Indpsolver
Yesterday at 10:58 AM
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Inequalities
sqing   0
Mar 21, 2025
Let $ a,b,c\geq 0 $ and $ a+b+c\geq 2+abc . $ Prove that
$$a^2+b^2+c^2- \frac{2}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{48}{25}$$$$a^2+b^2+c^2- \frac{3}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{91}{50}$$$$a^2+b^2+c^2- \frac{4}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{42}{25}$$$$a^2+b^2+c^2- \frac{8}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{3(7\sqrt{21}-27)}{25}$$$$a^2+b^2+c^2- \frac{9}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{8}{261+41 \sqrt{41}}$$
0 replies
sqing
Mar 21, 2025
0 replies
J
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Inequalities
sqing   29
N Mar 21, 2025 by SomeonecoolLovesMaths
Let $ a,b>0 $ and $ \frac{1}{a}+\frac{1}{b}=1. $ Prove that
$$(a^2-a+1)(b^2-b+1) \geq 9$$$$ (a^2-a+b+1)(b^2-b+a+1) \geq 25$$Let $ a,b>0 $ and $ \frac{1}{a}+\frac{1}{b}=\frac{2}{3}. $ Prove that
$$(a+8)(a^2-a+b+2)(b^2-b+5)\geq1331$$$$(a+10)(a^2-a+b+4)(b^2-b+7)\geq2197$$
29 replies
sqing
Mar 10, 2025
SomeonecoolLovesMaths
Mar 21, 2025
J
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Inequalities
sqing   12
N Mar 21, 2025 by sqing
Let $ a,b $ be real numbers such that $ a + b \geq |ab + 1|. $ Prove that$$ a^3 + b^3 \geq |a^3 b^3 + 1|$$Let $ a,b $ be real numbers such that $ 2(a + b ) \geq |ab + 1|. $ Prove that$$26( a^3 + b^3) \geq |a^3 b^3 + 1|$$Let $ a,b $ be real numbers such that $ 4(a + b) \geq 3|ab + 1|. $ Prove that$$148(a^3 + b^3) \geq27 |a^3 b^3 + 1|$$
12 replies
sqing
Mar 8, 2025
sqing
Mar 21, 2025
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a+b+c=3 ine
jokehim   4
N Mar 21, 2025 by lbh_qys
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that $$\color{black}{\frac{a\left(b+c\right)}{bc+3}+\frac{b\left(c+a\right)}{ca+3}+\frac{c\left(a+b\right)}{ab+3}\le \frac{3}{2}.}$$Proposed by Phan Ngoc Chau
4 replies
jokehim
Mar 18, 2025
lbh_qys
Mar 21, 2025
J
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Hard inequality
JK1603JK   2
N Mar 21, 2025 by lbh_qys
Let a,b,c>=0: ab+bc+ca=2 then find the minimum value P=\frac{a+b+c-2}{a^2b+b^2c+c^2a}
2 replies
JK1603JK
Mar 21, 2025
lbh_qys
Mar 21, 2025
J
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J
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A nice equality
hien   2
N Dec 9, 2005 by darij grinberg
let a, b, c are positive numbers. Prove that:
[code]\frac{a^{2}}{b^{2}+c^{2}}+frac{b^{2}}{c^{2}+a^{2}}+frac{c^{2}}{a^{2}+b^{2}}\gre\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}[/code]
2 replies
hien
Dec 9, 2005
darij grinberg
Dec 9, 2005
J
A nice equality
G H J
Reveal topic tags
algebra unsolved
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hien
645 posts
hien
#1 h  Dec 9, 2005, 8:08 AM • 2 Y
Y by Adventure10, Mango247
let a, b, c are positive numbers. Prove that:
\frac{a^{2}}{b^{2}+c^{2}}+frac{b^{2}}{c^{2}+a^{2}}+frac{c^{2}}{a^{2}+b^{2}}\gre\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}
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Virgil Nicula
7054 posts
Virgil Nicula
#2 h  Dec 9, 2005, 9:15 AM • 3 Y
Y by Adventure10, Adventure10, Mango247
$\{a,b,c\}\subset (0,\infty)\Longrightarrow \sum \frac{a^2}{b^2+c^2}\ge \sum \frac{a}{b+c}.$
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darij grinberg
6555 posts
darij grinberg
#3 h  Dec 9, 2005, 1:57 PM • 2 Y
Y by Adventure10, Mango247
In http://www.mathlinks.ro/Forum/viewtopic.php?t=50773 , I wrote:
darij grinberg wrote:
See http://www.mathlinks.ro/Forum/viewtopic.php?t=17207 and http://www.mathlinks.ro/Forum/viewtopic.php?t=34563 and the generalizations at http://www.mathlinks.ro/Forum/viewtopic.php?t=17230 , http://www.mathlinks.ro/Forum/viewtopic.php?t=23416 , http://www.mathlinks.ro/Forum/viewtopic.php?t=23752 .

darij
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