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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Find the number of integral solutions
Mathlover08092002   1
N 6 minutes ago by quasar_lord
Source: MTRP 2019 Class 11-Multiple Choice Question: Problem 2 :-
What is the number of integral solutions of the equation $a^{b^2}=b^{2a}$, where a > 0 and $|b|>|a|$
[list=1]
[*] 3
[*] 4
[*] 6
[*] 8
[/list]
1 reply
Mathlover08092002
Apr 9, 2020
quasar_lord
6 minutes ago
What can you say about f(x)
Mathlover08092002   3
N 10 minutes ago by quasar_lord
Source: MTRP 2019 Class 11-Multiple Choice Question: Problem 1 :-
Let $f : (0, \infty) \to \mathbb{R}$ is differentiable such that $\lim \limits_{x \to \infty} f(x)=2019$ Then which of the following is correct?
[list=1]
[*] $\lim \limits_{x \to \infty} f'(x)$ always exists but not necessarily zero.
[*] $\lim \limits_{x \to \infty} f'(x)$ always exists and is equal to zero.
[*] $\lim \limits_{x \to \infty} f'(x)$ may not exist.
[*] $\lim \limits_{x \to \infty} f'(x)$ exists if $f$ is twice differentiable.
[/list]
3 replies
Mathlover08092002
Apr 9, 2020
quasar_lord
10 minutes ago
FE on Stems
mathscrazy   4
N 18 minutes ago by SatisfiedMagma
Source: STEMS 2025 Category B4, C3
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all $x,y\in \mathbb{R}$, \[xf(y+x)+(y+x)f(y)=f(x^2+y^2)+2f(xy)\]Proposed by Aritra Mondal
4 replies
mathscrazy
Dec 29, 2024
SatisfiedMagma
18 minutes ago
xyz(x+y+z)=4
RnstTrjyn   5
N 19 minutes ago by sqing
Let $x, y, z$ be positive real numbers such that $xyz(x + y + z) = 4$. Prove that
$(x+y)^2+3(z+y)^2+(x+z)^2 \geq 8\sqrt7$
5 replies
1 viewing
RnstTrjyn
Feb 3, 2019
sqing
19 minutes ago
One inequality 1
prof.   2
N 20 minutes ago by invisibleman
If $x>0$ prove inequality $$\sqrt{x}\cdot (x+1)+x\cdot (x-4)+1\ge0.$$
2 replies
prof.
3 hours ago
invisibleman
20 minutes ago
Inequality
JK1603JK   1
N 20 minutes ago by CHESSR1DER
Source: unknown
Let a,b,c>=0 and ab+bc+ca>0 then prove \sqrt{a+b}+\sqrt{c+b}+\sqrt{a+c}\ge 2\sqrt[4]{ab+bc+ca}+\sqrt{\frac{a(b-c)^2+b(c-a)^2+c(a-b)^2}{ab+bc+ca}}
1 reply
1 viewing
JK1603JK
Today at 2:58 AM
CHESSR1DER
20 minutes ago
geometry coordinates
CHESSR1DER   0
29 minutes ago
Source: simplified version of Belarus TST
Points $A, B, C$ with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points $D, E, F$ with integer coordinates such that $AB = DE$, $AC = DF$, $BC = EF$
0 replies
CHESSR1DER
29 minutes ago
0 replies
Cutting a big square into smaller squares
nAalniaOMliO   5
N 32 minutes ago by RagvaloD
Source: Belarusian National Olympiad 2020
A $20 \times 20$ checkered board is cut into several squares with integer side length. The size of a square is it's side length.
What is the maximum amount of different sizes this squares can have?
5 replies
nAalniaOMliO
Jan 29, 2025
RagvaloD
32 minutes ago
Ah yes, very interesting
Quidditch   23
N 40 minutes ago by quantam13
Source: EGMO 2024 P4
For a sequence $a_1<a_2<\cdots<a_n$ of integers, a pair $(a_i,a_j)$ with $1\leq i<j\leq n$ is called interesting if there exists a pair $(a_k,a_l)$ of integers with $1\leq k<l\leq n$ such that $$\frac{a_l-a_k}{a_j-a_i}=2.$$For each $n\geq 3$, find the largest possible number of interesting pairs in a sequence of length $n$.
23 replies
Quidditch
Apr 14, 2024
quantam13
40 minutes ago
Nice FE as the First Day Finale
swynca   3
N 43 minutes ago by PerfectPlayer
Source: 2025 Turkey TST P3
Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x,y \in \mathbb{R}-\{0\}$,
$$ f(x) \neq 0 \text{ and } \frac{f(x)}{f(y)} + \frac{f(y)}{f(x)} - f \left( \frac{x}{y}-\frac{y}{x} \right) =2 $$
3 replies
swynca
Mar 18, 2025
PerfectPlayer
43 minutes ago
keep one card and discard the other
Scilyse   1
N an hour ago by g0USinsane777
Source: CGMO 2024 P2
There are $8$ cards on which the numbers $1$, $2$, $\dots$, $8$ are written respectively. Alice and Bob play the following game: in each turn, Alice gives two cards to Bob, who must keep one card and discard the other. The game proceeds for four turns in total; in the first two turns, Bob cannot keep both of the cards with the larger numbers, and in the last two turns, Bob also cannot keep both of the cards with the larger numbers. Let $S$ be the sum of the numbers written on the cards that Bob keeps. Find the greatest positive integer $N$ for which Bob can guarantee that $S$ is at least $N$.
1 reply
Scilyse
Jan 28, 2025
g0USinsane777
an hour ago
Simultaneous eqs. with matrix
RenheMiResembleRice   7
N an hour ago by RenheMiResembleRice
Source: Ningyi Hou
Solve the attached with steps
7 replies
RenheMiResembleRice
4 hours ago
RenheMiResembleRice
an hour ago
Find the minimum
sqing   6
N an hour ago by sqing
Source: 2019 China Mathematical Olympiad Hope League Summer Camp
Let $x,y,z $ be positive real number such that $xyz(x+y+z)=4.$ Find the minimum value of $(x+y)^2+2(y+z)^2+3(z+x)^2.$
6 replies
1 viewing
sqing
Aug 10, 2019
sqing
an hour ago
Interesting inequality
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b\geq 0  . $ Prove that
$$ a^4+b^4 +kab\geq\left(\sqrt{k(k+2)}-k\right)ab(a+b+k)$$Where $ k>0 . $
$$ a^4+b^4 +ab\geq (\sqrt 3-1)ab(a+b+1)$$$$ a^4+b^4 +2ab\geq 2(\sqrt 2-1)ab(a+b+2)$$
2 replies
sqing
5 hours ago
sqing
2 hours ago
Generalization of Janous'one
sqing   10
N Nov 9, 2020 by sqing
Source: Zhanyanzong
Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be positive numbers such that $a_1\geq a_2+a_3+\cdots+a_k $ $(k\geq 2).$ Prove that$$\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\geq n^2+(k-2)^2.$$
10 replies
sqing
Nov 3, 2020
sqing
Nov 9, 2020
Generalization of Janous'one
G H J
Source: Zhanyanzong
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sqing
41118 posts
#1
Y by
Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be positive numbers such that $a_1\geq a_2+a_3+\cdots+a_k $ $(k\geq 2).$ Prove that$$\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\geq n^2+(k-2)^2.$$
Z K Y
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Auditor
108 posts
#2
Y by
sqing wrote:
Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be positive numbers such that $a_1\geq a_2+a_3+\cdots+a_k $ $(k\geq 2).$ Prove that$$\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\geq n^2+(k-2)^2.$$

Sorry, but $k $ and $n $ depend to each other?
Z K Y
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sqing
41118 posts
#3
Y by
Auditor wrote:
sqing wrote:
Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be positive numbers such that $a_1\geq a_2+a_3+\cdots+a_k $ $(k\geq 2).$ Prove that$$\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\geq n^2+(k-2)^2.$$

Sorry, but $k $ and $n $ depend to each other?
Yeah.
Therefore , it is very nice.

Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c.$ Proof that$$\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 17 .$$Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c+d.$ Proof that$$\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 20 .$$
This post has been edited 1 time. Last edited by sqing, Nov 3, 2020, 8:04 AM
Z K Y
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sqing
41118 posts
#5
Y by
themathematicalmantra wrote:
yes i find this inequality quite fascinating. Came across such elegant problem in a while. Will solve and send solution
Thank you very much.
Z K Y
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richrow12
411 posts
#6
Y by
Just use Cauchy-Schwarz:
$$
((a_1+a_2+\ldots+a_k)+(a_{k+1}+\ldots+a_n))\left(\left(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_k}\right)+\frac{1}{a_{k+1}}+\ldots+\frac{1}{a_n}\right)\geq\left(\sqrt{(a_1+a_2+\ldots+a_k)\left(\frac{1}{a_1}+\frac{1}{a_2}\ldots+\frac{1}{a_k}\right)}+n-k\right)^2.
$$Now, denote $s=a_2+\ldots+a_k$ and $a=a_1$. Again, due to Cauchy-Schwarz (or AM-HM) we have
$$
\frac{1}{a_1}+\frac{1}{a_2}\ldots+\frac{1}{a_k}\geq\frac{1}{a}+\frac{(k-1)^2}{s}.
$$Thus, it remains to prove that
$$
\left(\sqrt{(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)}+n-k\right)^2\geq n^2+(k-2)^2.
$$Consider the expression
$$
(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)=(1+(k-1)^2)+\frac{s}{a}+(k-1)^2\cdot\frac{a}{s}.
$$Since $a\geq s$ and function $f(t)=\frac{1}{t}+(k-1)^2t$ is increasing on $[1+\infty)$ (due to $k\geq 2$) we have
$$
\left(\sqrt{(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)}+n-k\right)^2\geq \left(\sqrt{2(k-1)^2+2}+n-k\right)^2.
$$Therefore it suffices to prove that
$$
\left(\sqrt{2(k-1)^2+2}+n-k\right)^2\geq n^2+(k-2)^2
$$or
$$
2n\cdot\left(\sqrt{2(k-1)^2+2}-k\right)+\left(\sqrt{2(k-1)^2+2}-k\right)^2\geq (k-2)^2,
$$so since $\sqrt{2(k-1)^2+2}-k\geq 0$ we just need to check this inequality when $n=k$ (and it's easier to check initial inequality) or
$$
2(k-1)^2+2\geq k^2+(k-2)^2,
$$which is actually an equality.

Note. This solution isn't really artificial and is quite straightforward. Some motivation: firstly, we don't really care about $a_{k+1},\ldots,a_n$, so we can optimize the left hand for variables $a_{k+1},\ldots,a_n$ (so the first inequality makes all $a_{k+1},\ldots,a_n$ equal to 1). Secondly, we have restriction only on the sum $a_1+\ldots+a_k$, so we try to optimize left hand for $a_2,\ldots,a_k$ while the sum $a_2+\ldots+a_k$ is fixes (so our second inequality makes all variables $a_{2},\ldots,a_{k+1}$ equal). Basically, at this point we've reduced our problem to the case $a_{k+1}=\ldots=a_n=1$ and $a_2=\ldots=a_{k}=\frac{s}{k-1}$ with the only resrtriction $a_1\geq s$. Finally, the left hand now is a function of $t=a/s$, so we can find it's minimum on $[1+\infty)$ which is attained at $t=1$. After that it remains to check simple inequality of variables $n$ and $k$ (and again we can apply our "optimizing principle" since $n\geq k$ and $LHS-RHS$ is an increasing function of $n$).
This post has been edited 2 times. Last edited by richrow12, Nov 3, 2020, 1:48 PM
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Tintarn
9027 posts
#7
Y by
Solution
This post has been edited 2 times. Last edited by Tintarn, Nov 3, 2020, 9:51 AM
Z K Y
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sqing
41118 posts
#8
Y by
sqing wrote:
Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c.$ Proof that$$\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 17 .$$
Let $a,b,c$ and $d$ be positive real numbers such that $a \geq b+c+d.$ Proof that$$\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq 17 .$$
Thank you very much.
This post has been edited 1 time. Last edited by sqing, Jul 20, 2022, 2:40 PM
Z K Y
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Auditor
108 posts
#9
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richrow12 wrote:
Just use Cauchy-Schwarz:
$$
((a_1+a_2+\ldots+a_k)+(a_{k+1}+\ldots+a_n))\left(\left(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_k}\right)+\frac{1}{a_{k+1}}+\ldots+\frac{1}{a_n}\right)\geq\left(\sqrt{(a_1+a_2+\ldots+a_k)\left(\frac{1}{a_1}+\frac{1}{a_2}\ldots+\frac{1}{a_k}\right)}+n-k\right)^2.
$$Now, denote $s=a_2+\ldots+a_k$ and $a=a_1$. Again, due to Cauchy-Schwarz (or AM-HM) we have
$$
\frac{1}{a_1}+\frac{1}{a_2}\ldots+\frac{1}{a_k}\geq\frac{1}{a}+\frac{(k-1)^2}{s}.
$$Thus, it remains to prove that
$$
\left(\sqrt{(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)}+n-k\right)^2\geq n^2+(k-2)^2.
$$Consider the expression
$$
(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)=(1+(k-1)^2)+\frac{s}{a}+(k-1)^2\cdot\frac{a}{s}.
$$Since $a\geq s$ and function $f(t)=\frac{1}{t}+(k-1)^2t$ is increasing on $[1+\infty)$ (due to $k\geq 2$) we have
$$
\left(\sqrt{(a+s)\left(\frac{1}{a}+\frac{(k-1)^2}{s}\right)}+n-k\right)^2\geq \left(\sqrt{2(k-1)^2+2}+n-k\right)^2.
$$Therefore it suffices to prove that
$$
\left(\sqrt{2(k-1)^2+2}+n-k\right)^2\geq n^2+(k-2)^2
$$or
$$
2n\cdot\left(\sqrt{2(k-1)^2+2}-k\right)+\left(\sqrt{2(k-1)^2+2}-k\right)^2\geq (k-2)^2,
$$so since $\sqrt{2(k-1)^2+2}-k\geq 0$ we just need to check this inequality when $n=k$ (and it's easier to check initial inequality) or
$$
2(k-1)^2+2\geq k^2+(k-2)^2,
$$which is actually an equality.

Note. This solution isn't really artificial and is quite straightforward. Some motivation: firstly, we don't really care about $a_{k+1},\ldots,a_n$, so we can optimize the left hand for variables $a_{k+1},\ldots,a_n$ (so the first inequality makes all $a_{k+1},\ldots,a_n$ equal to 1). Secondly, we have restriction only on the sum $a_1+\ldots+a_k$, so we try to optimize left hand for $a_2,\ldots,a_k$ while the sum $a_2+\ldots+a_k$ is fixes (so our second inequality makes all variables $a_{2},\ldots,a_{k+1}$ equal). Basically, at this point we've reduced our problem to the case $a_{k+1}=\ldots=a_n=1$ and $a_2=\ldots=a_{k}=\frac{s}{k-1}$ with the only resrtriction $a_1\geq s$. Finally, the left hand now is a function of $t=a/s$, so we can find it's minimum on $[1+\infty)$ which attains at $t=1$. After that it remains to check simple inequality of variables $n$ and $k$ (and again we can apply our "optimizing principle" since $n\geq k$ and $LHS-RHS$ is an increasing function of $n$).

But we don't know whether $n \ge k $ or not? Let me know if I am wrong.
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GorgonMathDota
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@above We are only given $a_1, \dots, a_n$. If $k > n$, then there's a variable $a_k$ which is not given.
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yds
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sqing wrote:
Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be positive numbers such that $a_1\geq a_2+a_3+\cdots+a_k $ $(k\geq 2).$ Prove that$$\left(a_1+a_2+\cdots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\geq n^2+(k-2)^2.$$

$\text{Dear sqing can you add"n variable"or "Multivariate inequality",you know we AOPS' retrieval function is poor}$
$\text{Thank you very much!}$
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sqing
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Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be positive numbers such that $a_1\geq a_2+a_3+\cdots+a_k $ $(n\geq k\geq 2).$ Prove that$$\left(a^m_1+a^m_2+\cdots+a^m_n\right)\left(\frac{1}{a^m_1}+\frac{1}{a^m_2}+\cdots+\frac{1}{a^m_n}\right)\geq n^2+\frac{\left((k-1)^{|m|}-1\right)^2}{(k-1)^{|m|-1}}.$$Where $m\in R.$
This post has been edited 3 times. Last edited by sqing, Nov 9, 2020, 5:27 AM
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