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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 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
Number theory
MathsII-enjoy   1
N 2 minutes ago by MathsII-enjoy
Prove that when $x^p+y^p$ | $(p^2-1)^n$ with $x,y$ are positive integers and $p$ is prime ($p>3$), we get: $x=y$
1 reply
MathsII-enjoy
Yesterday at 3:22 PM
MathsII-enjoy
2 minutes ago
(not so) small set of residues generates all of F_p upon applying Q many times
62861   14
N 7 minutes ago by john0512
Source: RMM 2019 Problem 6
Find all pairs of integers $(c, d)$, both greater than 1, such that the following holds:

For any monic polynomial $Q$ of degree $d$ with integer coefficients and for any prime $p > c(2c+1)$, there exists a set $S$ of at most $\big(\tfrac{2c-1}{2c+1}\big)p$ integers, such that
\[\bigcup_{s \in S} \{s,\; Q(s),\; Q(Q(s)),\; Q(Q(Q(s))),\; \dots\}\]contains a complete residue system modulo $p$ (i.e., intersects with every residue class modulo $p$).
14 replies
62861
Feb 24, 2019
john0512
7 minutes ago
find positive n so that exists prime p with p^n-(p-1)^n$ a power of 3
parmenides51   13
N 7 minutes ago by SimplisticFormulas
Source: JBMO Shortlist 2017 NT5
Find all positive integers $n$ such that there exists a prime number $p$, such that $p^n-(p-1)^n$ is a power of $3$.

Note. A power of $3$ is a number of the form $3^a$ where $a$ is a positive integer.
13 replies
parmenides51
Jul 25, 2018
SimplisticFormulas
7 minutes ago
Functional equation of nonzero reals
proglote   5
N 9 minutes ago by TheHimMan
Source: Brazil MO 2013, problem #3
Find all injective functions $f\colon \mathbb{R}^* \to \mathbb{R}^* $ from the non-zero reals to the non-zero reals, such that \[f(x+y) \left(f(x) + f(y)\right) = f(xy)\] for all non-zero reals $x, y$ such that $x+y \neq 0$.
5 replies
proglote
Oct 24, 2013
TheHimMan
9 minutes ago
5-th powers is a no-go - JBMO Shortlist
WakeUp   8
N 13 minutes ago by sansgankrsngupta
Prove that there are are no positive integers $x$ and $y$ such that $x^5+y^5+1=(x+2)^5+(y-3)^5$.

Note
8 replies
WakeUp
Oct 30, 2010
sansgankrsngupta
13 minutes ago
IMO Genre Predictions
ohiorizzler1434   54
N 15 minutes ago by sansgankrsngupta
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
54 replies
ohiorizzler1434
May 3, 2025
sansgankrsngupta
15 minutes ago
F.E....can you solve it?
Jackson0423   15
N 16 minutes ago by MathsII-enjoy
Find all functions \( f : \mathbb{R} \to \mathbb{R} \) such that
\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]for all real numbers \( x \) satisfying \( f(x) \neq 1 \).
15 replies
+1 w
Jackson0423
Yesterday at 1:27 PM
MathsII-enjoy
16 minutes ago
Chess game challenge
adihaya   20
N 16 minutes ago by cursed_tangent1434
Source: 2014 BAMO-12 #5
A chess tournament took place between $2n+1$ players. Every player played every other player once, with no draws. In addition, each player had a numerical rating before the tournament began, with no two players having equal ratings. It turns out there were exactly $k$ games in which the lower-rated player beat the higher-rated player. Prove that there is some player who won no less than $n-\sqrt{2k}$ and no more than $n+\sqrt{2k}$ games.
20 replies
adihaya
Feb 22, 2016
cursed_tangent1434
16 minutes ago
Permutations inequality
OronSH   13
N 17 minutes ago by sansgankrsngupta
Source: ISL 2023 A5
Let $a_1,a_2,\dots,a_{2023}$ be positive integers such that
[list=disc]
[*] $a_1,a_2,\dots,a_{2023}$ is a permutation of $1,2,\dots,2023$, and
[*] $|a_1-a_2|,|a_2-a_3|,\dots,|a_{2022}-a_{2023}|$ is a permutation of $1,2,\dots,2022$.
[/list]
Prove that $\max(a_1,a_{2023})\ge 507$.
13 replies
OronSH
Jul 17, 2024
sansgankrsngupta
17 minutes ago
Find all real numbers
sqing   5
N 18 minutes ago by ytChen
Source: IMOC 2021 A1
Find all real numbers x that satisfies$$\sqrt{\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}-\frac{1}{\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}}+\sqrt{1-\frac{1}{\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}}=x.$$2021 IMOC Problems
5 replies
sqing
Aug 11, 2021
ytChen
18 minutes ago
4-var inequality
sqing   0
22 minutes ago
Source: SXTB (4)2025 Q2837
Let $ a,b,c,d> 0  $. Prove that
$$   \frac{1}{(3a+1)^4}+ \frac{1}{(3b+1)^4}+\frac{1}{(3c+1)^4}+\frac{1}{(3d+1)^4} \geq \frac{1}{16(3abcd+1)}$$
0 replies
sqing
22 minutes ago
0 replies
Inspired by Bet667
sqing   0
35 minutes ago
Source: Own
Let $ a,b $ be a real numbers such that $a^2+kab+b^2\ge a^3+b^3.$Prove that$$a+b\leq k+2$$Where $ k\geq 0. $
0 replies
sqing
35 minutes ago
0 replies
BMO 2024 SL A4
MuradSafarli   3
N an hour ago by sqing
A4.
Let \(a \geq b \geq c \geq 0\) be real numbers such that \(ab + bc + ca = 3\).
Prove that:
\[
3 + (2 - \sqrt{3}) \cdot \frac{(b-c)^2}{b+(\sqrt{3}-1)c} \leq a+b+c
\]and determine all the cases when the equality occurs.
3 replies
MuradSafarli
Apr 27, 2025
sqing
an hour ago
an exponential inequality with two variables
teresafang   5
N an hour ago by teresafang
x and y are positive real numbers.prove that [(x^y)/y]^(1/2)+[(y^x)/x]^(1/2)>=2.
sorry.I’m not good at English.Also I don’t know how to use Letax.
5 replies
teresafang
May 4, 2025
teresafang
an hour ago
The prime inequality learning problem
orl   138
N Apr 25, 2025 by Ilikeminecraft
Source: IMO 1995, Problem 2, Day 1, IMO Shortlist 1995, A1
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that
\[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}.
\]
138 replies
orl
Nov 9, 2005
Ilikeminecraft
Apr 25, 2025
The prime inequality learning problem
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 1995, Problem 2, Day 1, IMO Shortlist 1995, A1
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Scilyse
387 posts
#138
Y by
Here
\begin{align*}
    \sum \frac{1}{a^3 (b + c)} &= \sum \frac{\frac{1}{a^2}}{ab + ac} \\
    &\geq \frac{(\sum \frac{1}{a})^2}{2\sum ab}\text{ (by Titu)} \\
    &= \frac{(\sum ab)^2}{2a^2 b^2 c^2 \sum ab} \\
    &= \frac{\sum ab}{2a^2 b^2 c^2} \\
    &= \frac{1}{2} \sum ab \\
    &\geq \frac{3}{2}\text{ (by AM-GM)}
\end{align*}as desired.
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Jishnu4414l
154 posts
#139
Y by
$=\sum_{cyc}\frac{b^2c^2}{a(b+c)} \geq \frac{ab+bc+ca}{2} \geq \frac{3}{2}$ by Titu's lemma and AM-GM respectively.
We are thus done.
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Math4Life7
1703 posts
#140
Y by
By Holder's inequality we get \[\left(\sum_{\text{cyc}} a(b+c)\right)^{\frac{1}{2}} \left(\sum_{\text{cyc}} \frac{1}{a^3(b+c)} \right)^{\frac{1}{2}} \geq \left(\frac 1a + \frac 1b + \frac 1c \right) = (ab+ac+bc)\]Thus it remains to prove that $\frac{ab+ac+bc}{2} \geq \frac{3}{2}$. This is obvious from Muirhead with $\left(\frac 43, \frac 43, \frac 13 \right) \succ (1, 1, 1)$
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kuzeyaloglu
16 posts
#142
Y by
$a=b=c$ works
Multiplying by the denominators and using $abc = 1$ gives
$$\sum{a^3b^2 a^2b^3+ bc + b^4c^4} \ge 3 + \frac{3}{2}\sum{a^2b + b^2a}$$From Muirhead's Inequality we have
$$\sum{a^2b + b^2a} = \sum{a^{\frac{8}{3}} b^{\frac{5}{3}} c^{\frac{2}{3}}} \le \sum{a^3b^2 + a^2b^3} $$$$\sum{a^2b + b^2a} = \sum{a^{\frac{11}{3}} b^{\frac{8}{3}} c^{\frac{5}{3}}} \le \sum{a^4b^4} $$$$ab+ac+bc \ge 3$$Adding them all together gives the desired inequality.
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SilverBlaze_SY
66 posts
#143
Y by
$$\frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}$$As $abc=1, $Multiplying $(abc)^2$ on the $LHS$, we get:
$$\frac{(bc)^2}{a(b+c)} + \frac{(ca)^2}{b(c+a)}+\frac{(ab^2)}{c(a+b)} \geq \frac{(ab+bc+ca)^2}{a(b+c)+b(c+a)+c(a+b)}=\frac{(ab+bc+ca)}{2}$$[By Titu's Lemma]

Again, $\frac{ab+bc+ca}{2} \geq \frac{3 \sqrt[3]{a^2b^2c^2}}{2} = \frac{3}{2}$ by AM-GM. Hence, we're done!
This post has been edited 1 time. Last edited by SilverBlaze_SY, Mar 17, 2024, 6:21 AM
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anduran
481 posts
#144
Y by
a variation
If $a+b+c=3,$ then
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \geq \frac{3}{2}.$$
This post has been edited 1 time. Last edited by anduran, Mar 17, 2024, 6:31 AM
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D_S
105 posts
#145
Y by
anduran wrote:
a variation
If $a+b+c=3,$ then
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \geq \frac{3}{2}.$$

Write$b+c = 3-a$ etc, and use Jensen on $f(x) = \frac{1}{x^3(3-x)}$.
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dudade
139 posts
#146
Y by
Notice that
\begin{align*}
\frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)} &= \frac {a^2b^2c^2}{a^{3}\left(b + c\right)} + \frac {a^2b^2c^2}{b^{3}\left(c + a\right)} + \frac {a^2b^2c^2}{c^{3}\left(a + b\right)} \\
&= \frac {b^2c^2}{ab+ac} + \frac {c^2a^2}{bc+ba} + \frac {a^2b^2}{ca+cb}.
\end{align*}By Titu's Lemma,
\begin{align*}
\frac {b^2c^2}{ab+ac} + \frac {c^2a^2}{bc+ba} + \frac {a^2b^2}{ca+cb} \geq \dfrac{(ab+bc+ca)^2}{2(ab+bc+ca)} = \dfrac{ab+bc+ca}{2}.
\end{align*}By AM-GM, note that $ab + bc + ca \geq 3\sqrt[3]{a^2b^2c^2} = 3$, as desired.
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newton_24
77 posts
#148
Y by
I was to lazy to read all the solutions so never mind if the solution is same.
By using titu's lemma
$\frac{\frac{1^2}{a^2}}{a(b+c)} + \frac{\frac{1^2}{b^2}}{b(a+c)}+\frac{\frac{1^2}{c^2}}{c(b+a)} \ge \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{2(ab+bc+ac)}$
After some simple calculations we finally get:
$\frac{\frac{1^2}{a^2}}{a(b+c)} + \frac{\frac{1^2}{b^2}}{b(a+c)}+\frac{\frac{1^2}{c^2}}{c(b+a)} \ge \frac{ab+bc+ac}{2}$
Using A.M.-G.M. inequality on $ \frac{ab+bc+ac}{2}$
$\frac{ab+bc+ac}{2*3} \ge \frac{\sqrt[3]{(abc)^2)}}{2}$
hence,$\frac{\frac{1^2}{a^2}}{a(b+c)} + \frac{\frac{1^2}{b^2}}{b(a+c)}+\frac{\frac{1^2}{c^2}}{c(b+a)} \ge \frac{ab+bc+ac}{2}\ge \frac{\sqrt[3]{(abc)^2}*3}{2}\ge \frac{3}{2}$
This post has been edited 1 time. Last edited by newton_24, Oct 2, 2024, 5:09 PM
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little-fermat
147 posts
#149
Y by
I discussed this problem on my youtube channel (little fermat) Video in my inequalities tutorial playlist.
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RetroFuel
11 posts
#150
Y by
Substitute $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$. We get,
$$\sum_{cyc} \frac{y^3z}{x^3y+x^2z^2}$$
Multiply and divide by $yz$
$$\sum_{cyc} \frac{y^4z^2}{x^3y^2z+x^2yz^3} \geq  \frac{(x^2y+y^2z+z^2x)^2}{2xyz(x^2y+y^2z+z^2x)} = \frac{x^2y+y^2z+z^2x}{2xyz}$$This inequality is by Cauchy-Schwarz. It is sufficient to prove that
$$\frac{x^2y+y^2z+z^2x}{2xyz} \geq \frac{3}{2}$$$$\implies x^2y+y^2z+z^2x \geq 3xyz $$
This is true by AM-GM
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Sadigly
159 posts
#151
Y by
$\sum\frac{1}{a^3(b+c)}=\sum\frac{\frac1{a^2}}{ab+ac}\geq\frac{(\frac1{a}+\frac1{b}+\frac1{c})^2}{2(ab+ac+bc)}=\frac{(ab+bc+ca)^2}{2(ab+ac+bc)}=\frac{(ab+bc+ca)}{2}\geq\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac32$
This post has been edited 3 times. Last edited by Sadigly, Dec 16, 2024, 7:48 PM
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cappucher
94 posts
#153
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First, let $x = \frac{1}{a}$, $y = \frac{1}{b}$, and $z = \frac{1}{c}$. Our sum is then

\[\sum_{\text{cyc}} \frac{1}{\frac{1}{x^3}\left(\frac{1}{y} + \frac{1}{z} \right)} = \sum_{\text{cyc}} \frac{1}{\frac{y + z}{x^3yz}} = \sum_{\text{cyc}} \frac{1}{\frac{y + z}{x^2}} = \sum_{\text{cyc}} \frac{x^2}{y+z}\]
Applying Titu and AM-GM, we have

\[\sum_{\text{cyc}} \frac{x^2}{y+z} \geq \frac{(x+y+z)^2}{2(x+y+z)} = \frac{x+y+z}{2} \geq \frac{3\sqrt[3]{xyz}}{2} = \frac{3}{2}\]
as desired.
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Marcus_Zhang
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#154
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Cool
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Ilikeminecraft
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#155
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Cauchy gives:
\begin{align*}
  \left(\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)}\right)(a(b + c) + b(a + c) + c(a + b)) & \geq 9\left(\frac1a + \frac1b + \frac1c\right)^2
\end{align*}Hence, all that remains is to prove that $ab + ac + bc \geq 3.$ However, by AM-GM, this is trivial.

For equality, note that in the last step, we used AM-GM. Thus, we have that $ab = ac = bc,$ and so $a = b = c,$ and hence $a = b = c = 1$
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