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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
inequality
xytunghoanh   3
N 7 minutes ago by sqing
For $a,b,c\ge 0$. Let $a+b+c=3$.
Prove or disprove
\[\sum ab +\sum ab^2 \le 6\]
3 replies
xytunghoanh
2 hours ago
sqing
7 minutes ago
f(f(n))=2n+2
Jackson0423   1
N 17 minutes ago by jasperE3
Source: 2013 KMO Second Round

Let \( f : \mathbb{N} \to \mathbb{N} \) be a function satisfying the following conditions for all \( n \in \mathbb{N} \):
\[
\begin{cases}
f(n+1) > f(n) \\
f(f(n)) = 2n + 2
\end{cases}
\]Find the value of \( f(2013) \).
1 reply
Jackson0423
Yesterday at 4:07 PM
jasperE3
17 minutes ago
Proving ZA=ZB
nAalniaOMliO   8
N an hour ago by Mathgloggers
Source: Belarusian National Olympiad 2025
Point $H$ is the foot of the altitude from $A$ of triangle $ABC$. On the lines $AB$ and $AC$ points $X$ and $Y$ are marked such that the circumcircles of triangles $BXH$ and $CYH$ are tangent, call this circles $w_B$ and $w_C$ respectively. Tangent lines to circles $w_B$ and $w_C$ at $X$ and $Y$ intersect at $Z$.
Prove that $ZA=ZH$.
Vadzim Kamianetski
8 replies
nAalniaOMliO
Mar 28, 2025
Mathgloggers
an hour ago
Hard geometry
Lukariman   1
N an hour ago by Lukariman
Given circle (O) and chord AB with different diameters. The tangents of circle (O) at A and B intersect at point P. On the small arc AB, take point C so that triangle CAB is not isosceles. The lines CA and BP intersect at D, BC and AP intersect at E. Prove that the centers of the circles circumscribing triangles ACE, BCD and OPC are collinear.
1 reply
Lukariman
an hour ago
Lukariman
an hour ago
Incircle triangles inequality
MathMystic33   1
N an hour ago by Quantum-Phantom
Source: 2025 Macedonian Team Selection Test P5
Let $\triangle ABC$ be a triangle with side‐lengths $a,b,c$, incenter $I$, and circumradius $R$. Denote by $P$ the area of $\triangle ABC$, and let $P_1,\;P_2,\;P_3$ be the areas of triangles $\triangle ABI$, $\triangle BCI$, and $\triangle CAI$, respectively. Prove that
\[
\frac{abc}{12R}
\;\le\;
\frac{P_1^2 + P_2^2 + P_3^2}{P}
\;\le\;
\frac{3R^3}{4\sqrt[3]{abc}}.
\]
1 reply
MathMystic33
Yesterday at 6:06 PM
Quantum-Phantom
an hour ago
Concurrency of tangent touchpoint lines on thales circles
MathMystic33   2
N an hour ago by Diamond-jumper76
Source: 2024 Macedonian Team Selection Test P4
Let $\triangle ABC$ be an acute scalene triangle. Denote by $k_A$ the circle with diameter $BC$, and let $B_A,C_A$ be the contact points of the tangents from $A$ to $k_A$, chosen so that $B$ and $B_A$ lie on opposite sides of $AC$ and $C$ and $C_A$ lie on opposite sides of $AB$. Similarly, let $k_B$ be the circle with diameter $CA$, with tangents from $B$ touching at $C_B,A_B$, and $k_C$ the circle with diameter $AB$, with tangents from $C$ touching at $A_C,B_C$.
Prove that the lines $B_AC_A, C_BA_B, A_CB_C$ are concurrent.
2 replies
MathMystic33
Yesterday at 7:41 PM
Diamond-jumper76
an hour ago
Find all possible values of q-p
yunxiu   18
N 2 hours ago by Jupiterballs
Source: 2012 European Girls’ Mathematical Olympiad P5
The numbers $p$ and $q$ are prime and satisfy
\[\frac{p}{{p + 1}} + \frac{{q + 1}}{q} = \frac{{2n}}{{n + 2}}\]
for some positive integer $n$. Find all possible values of $q-p$.

Luxembourg (Pierre Haas)
18 replies
yunxiu
Apr 13, 2012
Jupiterballs
2 hours ago
Inspired by nhathhuyyp5c
sqing   3
N 2 hours ago by sqing
Source: Own
Let $ x,y>0, x+2y- 3xy\leq 4. $Prove that
$$ \frac{1}{x^2} + 2y^2 + 3x + \frac{4}{y} \geq 3\left(2+\sqrt[3]{\frac{9}{4} }\right)$$
3 replies
sqing
3 hours ago
sqing
2 hours ago
Easy but Nice 12
TelvCohl   1
N 2 hours ago by Luis González
Source: Own
Given a $ \triangle ABC $ with orthocenter $ H $ and a point $ P $ lying on the Euler line of $ \triangle ABC. $ Prove that the midpoint of $ PH $ lies on the Thomson cubic of the pedal triangle of $ P $ WRT $ \triangle ABC. $
1 reply
TelvCohl
Mar 8, 2025
Luis González
2 hours ago
Similar Problems
Saucepan_man02   2
N 2 hours ago by quasar_lord
Could anyone post some problems which are similar to the below problem:

Find the real solution of: $$x^9+9/8 x^6+27/64 x^3-x+219/512.$$
Sol(outline)
2 replies
Saucepan_man02
May 12, 2025
quasar_lord
2 hours ago
Inspired by old results
sqing   0
2 hours ago
Source: Own
Let $ a,b,c>0 $ . Prove that
$$\frac{a+kb}{b+c}+\frac{b+kc}{c+a}+\frac{c+ka}{a+b}\geq \frac{3(k+1)}{2}$$W here $-1 \leq k \leq  \frac{537}{90}.$
0 replies
sqing
2 hours ago
0 replies
orthocenter on sus circle
DVDTSB   2
N 2 hours ago by Diamond-jumper76
Source: Romania TST 2025 Day 2 P1
Let \( ABC \) be an acute triangle with \( AB < AC \), and let \( O \) be the center of its circumcircle. Let \( A' \) be the reflection of \( A \) with respect to \( BC \). The line through \( O \) parallel to \( BC \) intersects \( AC \) at \( F \), and the tangent at \( F \) to the circle \( \odot(BFC) \) intersects the line through \( A' \) parallel to \( BC \) at point \( M \). Let \( K \) be a point on the ray \( AB \), starting at \( A \), such that \( AK = 4AB \).
Show that the orthocenter of triangle \( ABC \) lies on the circle with diameter \( KM \).

Proposed by Radu Lecoiu

2 replies
DVDTSB
Yesterday at 12:18 PM
Diamond-jumper76
2 hours ago
problem 5
termas   74
N 2 hours ago by maromex
Source: IMO 2016
The equation
$$(x-1)(x-2)\cdots(x-2016)=(x-1)(x-2)\cdots (x-2016)$$is written on the board, with $2016$ linear factors on each side. What is the least possible value of $k$ for which it is possible to erase exactly $k$ of these $4032$ linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
74 replies
termas
Jul 12, 2016
maromex
2 hours ago
I think I know why this problem was rejected by IMO PSC several times...
mshtand1   1
N 3 hours ago by sarjinius
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 11.8
Exactly $102$ country leaders arrived at the IMO. At the final session, the IMO chairperson wants to introduce some changes to the regulations, which the leaders must approve. To pass the changes, the chairperson must gather at least \(\frac{2}{3}\) of the votes "FOR" out of the total number of leaders. Some leaders do not attend such meetings, and it is known that there will be exactly $81$ leaders present. The chairperson must seat them in a square-shaped conference hall of size \(9 \times 9\), where each leader will be seated in a designated \(1 \times 1\) cell. It is known that exactly $28$ of these $81$ leaders will surely support the chairperson, i.e., they will always vote "FOR." All others will vote as follows: At the last second of voting, they will look at how their neighbors voted up to that moment — neighbors are defined as leaders seated in adjacent cells \(1 \times 1\) (sharing a side). If the majority of neighbors voted "FOR," they will also vote "FOR." If there is no such majority, they will vote "AGAINST." For example, a leader seated in a corner of the hall has exactly $2$ neighbors and will vote "FOR" only if both of their neighbors voted "FOR."

(a) Can the IMO chairperson arrange their $28$ supporters so that they vote "FOR" in the first second of voting and thereby secure a "FOR" vote from at least \(\frac{2}{3}\) of all $102$ leaders?

(b) What is the maximum number of "FOR" votes the chairperson can obtain by seating their 28 supporters appropriately?

Proposed by Bogdan Rublov
1 reply
mshtand1
Mar 14, 2025
sarjinius
3 hours ago
International FE olympiad P3
Functional_equation   22
N Apr 28, 2025 by jasperE3
Source: IFEO Day 1 P3
Find all functions $f:\mathbb R^+\rightarrow \mathbb R^+$ such that$$f(f(x)f(f(x))+y)=xf(x)+f(y)$$for all $x,y\in \mathbb R^+$

$\textit{Proposed by Functional\_equation, Mr.C and TLP.39}$
22 replies
Functional_equation
Feb 6, 2021
jasperE3
Apr 28, 2025
International FE olympiad P3
G H J
G H BBookmark kLocked kLocked NReply
Source: IFEO Day 1 P3
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Functional_equation
530 posts
#1 • 4 Y
Y by Aritra12, Mathematicsislovely, ywq233, megarnie
Find all functions $f:\mathbb R^+\rightarrow \mathbb R^+$ such that$$f(f(x)f(f(x))+y)=xf(x)+f(y)$$for all $x,y\in \mathbb R^+$

$\textit{Proposed by Functional\_equation, Mr.C and TLP.39}$
This post has been edited 2 times. Last edited by Functional_equation, Feb 6, 2021, 6:15 AM
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Functional_equation
530 posts
#2 • 1 Y
Y by megarnie
Note: We will not share the official solutions yet. We will share when the Solution Bookled is ready. :)
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pco
23511 posts
#3 • 10 Y
Y by Pitagar, Functional_equation, Mathematicsislovely, EmilXM, Atpar, ywq233, Supercali, OlympusHero, megarnie, terg
Functional_equation wrote:
Find all functions $f:\mathbb R^+\rightarrow \mathbb R^+$ such that$$f(f(x)f(f(x))+y)=xf(x)+f(y)$$for all $x,y\in \mathbb R^+$
Let $P(x,y)$ be the assertion $f(f(x)f(f(x))+y)=xf(x)+f(y)$

Simple induction implies
New assertion $Q(x,y,n)$ : $f(y+nf(x)f(f(x)))=f(y)+nxf(x)$ $\forall x,y>0$, $\forall n\in\mathbb Z_{\ge 0}$

1) New assertion $R(x,y)$ : $f(y)\ge \frac x{f(f(x))}y-xf(x)$ $\forall x,y>0$
Proof

2) $f(f(x))=x\quad\forall x>0$ (and so $f(x)$ is bijective)
Proof

3) $\boxed{f(x)=x\quad\forall x>0}$
Proof
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Functional_equation
530 posts
#4
Y by
@PCO congratulations! :first:
Note: None of the participants in the contest was able to solve this problem.
Z K Y
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Functional_equation
530 posts
#5 • 3 Y
Y by ywq233, Aritra12, megarnie
This is IFEO SL A7
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Functional_equation
530 posts
#6 • 1 Y
Y by MeowX2
Official Solution
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JuniorPerelman
114 posts
#7 • 1 Y
Y by Mango247
I think that there is a simple method
Juste set x=0 and since f(0) exist we can also set y=-f(0)f(f(0))
Now we get f(0)=-f(0)f(f(0)) and then we have f(0)=0 or f(f(0))=-1
And the remainder follow from both cases
And we find that the only solution is f(x)=x
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JuniorPerelman
114 posts
#8 • 1 Y
Y by Mango247
Functional_equation wrote:
@PCO congratulations! :first:
Note: None of the participants in the contest was able to solve this problem.

Are u sure?
Cause it seems like obvious
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DottedCaculator
7354 posts
#9
Y by
Hello, $f(0)$ does not exist, as the domain of $f$ is all positive real numbers.
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JuniorPerelman
114 posts
#10 • 1 Y
Y by Mango247
DottedCaculator wrote:
Hello, $f(0)$ does not exist, as the domain of $f$ is all positive real numbers.

Why do you except 0 to positive real number?
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DottedCaculator
7354 posts
#11 • 1 Y
Y by Mango247
0 is not positive.
Z K Y
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JuniorPerelman
114 posts
#12 • 3 Y
Y by Mango247, Mango247, Mango247
DottedCaculator wrote:
0 is not positive.

Is it negative?
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phoenixfire
372 posts
#13 • 2 Y
Y by The_Musilm, megarnie
$\mathbb R^+$ does not contain a zero. Unless you define it differently.
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JuniorPerelman
114 posts
#14
Y by
JuniorPerelman wrote:
DottedCaculator wrote:
0 is not positive.

Is it negative?

I see know
Excuse me french maths and English maths are different
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rama1728
800 posts
#15
Y by
JuniorPerelman wrote:
JuniorPerelman wrote:
DottedCaculator wrote:
0 is not positive.

Is it negative?

I see know
Excuse me french maths and English maths are different

Wdym french maths and english maths?
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Bradygho
2507 posts
#16 • 1 Y
Y by rama1728
rama1728 wrote:
Wdym french maths and english maths?

I think JuniorPerelman means French math terms are differently defined than English math terms. Though the overall essence is the same.
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ZETA_in_olympiad
2211 posts
#17 • 2 Y
Y by rama1728, Mango247
JuniorPerelman wrote:
DottedCaculator wrote:
0 is not positive.

Is it negative?

It's non-negative and non-positive.
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MathLuis
1526 posts
#18
Y by
Let $P(x,y)$ the assertion of the given F.E.
Claim 1: $f$ is an involution.
Proof: By this lemma:
Lemma wrote:
If $f,g,h:\mathbb{R}^+ \to\mathbb{R}^+$ be functions such that
\[f(g(x)+y)=h(x)+f(y) \quad \forall x,y>0\]then $\frac{g}{h}$ is a constant function.
We have that $f(f(x))=cx$ so re-write the F.E. as
$$f(cxf(x)+y)=xf(x)+f(y)$$Now take $f$ in both sides of $P(f(x),y)$ and then use the result of $P(x,f(y))$ to get that
$$c^3xf(x)+cy=f(f(c^2xf(x)+y))=f(cxf(x)+f(y))=xf(x)+cy \implies c=1$$Hence $f(f(x))=x$ so $f$ is involutive.
Claim 2: $f(y)+xf(x) \ge y$ for all $x,y$
Proof: Assume that $y>f(y)+xf(x)$ for some $x,y$ then we have $\frac{y}{xf(x)}>\frac{f(y)}{xf(x)}+1$ now since for any positive real $r$ there exists a positive integer $n$ such that $r+1>n \ge r$ we have that there exists $n$ positive integer such that $\frac{y}{xf(x)}>\frac{f(y)}{xf(x)}+1>n \ge \frac{f(y)}{xf(x)}$ which means that $y>nxf(x) \ge f(y)$. Now by easy induction we get that
$$f(nxf(x)+y)=nxf(x)+f(y)$$And on this equation do $y=y-nxf(x)$ to get that
$$f(y-nxf(x))=f(y)-nxf(x) \le 0 \; \text{contradiction!!}$$Hence our claim is true.
Finishing: Call the assertion of Claim 2 $Q(x,y)$ so now by $Q(x,f(y))$ we get that $y+xf(x) \ge f(y)$ and multpliying by $y$ in both sides $y^2+yxf(x) \ge yf(y)$ and by letting $y \to 0$ we have that $yf(y)$ is as smaller as we want so on $Q(x,y)$ set $xf(x)$ as smaller as we want so we get $f(y) \ge y$ but by setting $y=f(y)$ we get $y \ge f(y)$ hence $f(y)=y$.
Hence $\boxed{f(x)=x \; \forall x \in \mathbb R^+}$ is a solution thus we are done :D
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ZETA_in_olympiad
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#19
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MathLuis wrote:
Lemma wrote:
If $f,g,h:\mathbb{R}^+ \to\mathbb{R}^+$ be functions such that
\[f(g(x)+y)=h(x)+f(y) \quad \forall x,y>0\]then $\frac{g}{h}$ is a constant function.

You have to prove the lemma (as it's not obvious nor trivial) if you can't then you should show where it's proved i.e \ cite where it can be found.
This post has been edited 2 times. Last edited by ZETA_in_olympiad, Jun 5, 2022, 9:32 AM
Reason: i.e
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megarnie
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#20 • 1 Y
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ZETA_in_olympiad wrote:
MathLuis wrote:
Lemma wrote:
If $f,g,h:\mathbb{R}^+ \to\mathbb{R}^+$ be functions such that
\[f(g(x)+y)=h(x)+f(y) \quad \forall x,y>0\]then $\frac{g}{h}$ is a constant function.

You have to prove the lemma (as it's not obvious nor trivial) if you can't then you should show where it's proved i.e \ cite where it can be found.

The lemma can be found https://artofproblemsolving.com/community/c6h2807267p24753821
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navi_09220114
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Solution with the Malaysian team:

The main idea (almost ubiquitous with R+ FEs) is to estimate $f(small)$ as negligible, and this is often done by induction on $\mathbb{Z}$, or by introducing a new variable - similar to IMOSL 2007 A4.

$\textbf{Part 1.}$ Observe that we have for all $m, n\in \mathbb{Z}$, $$f(mf(x)f(f(x))+nf(y)f(f(y))+z)=mxf(x)+nyf(y)+f(z)$$Let $t=mf(x)f(f(x))+nf(y)f(f(y))+z$ and vary $z>0$ (intuitively, this $z$ is small), this implies that $$t>mf(x)f(f(x))+nf(y)f(f(y)) \Rightarrow f(t)>mxf(x)+nyf(y)$$For simplicity let $a=f(x)f(f(x)), b=f(y)f(f(y)), c=xf(x), d=yf(y)$, so that we get $$t>ma+nb \Rightarrow f(t)>mc+nd$$If $\frac{a}{b}<\frac{c}{d}$ then let $n=-\left\lfloor\frac{ma}{b}\right\rfloor$ we get $$b+1>ma-\left\lfloor\frac{ma}{b}\right\rfloor b \Rightarrow f(b+1)>mc-\left\lfloor\frac{ma}{b}\right\rfloor d>m\left(c-\frac{a}{b} d\right) - d$$which is a contradiction by taking $m$ arbitrarily huge.

So $\frac{a}{b}\ge\frac{c}{d}$. By swapping $x$ and $y$, we get that $\frac{a}{b}=\frac{c}{d}$ must hold, that is $$\frac{f(f(x))}{x}=\frac{f(f(y))}{y}$$for all $x$ and $y$. Hence $f(f(x))=cx$ for some $c>0$.

$\textbf{Part 2.}$ The original equation becomes $$f(cxf(x)+y)=xf(x)+f(y)$$Consider $P(x,f(y))$ and $P(f(x),y)$, then $$f(cxf(x)+f(y))=xf(x)+cy$$$$f(c^2xf(x)+y)=cxf(x)+f(y)$$So this gives $$c^3xf(x)+cy=f(f(c^2xf(x)+y))=f(cxf(x)+f(y))=xf(x)+cy$$which gives $c=1$. So $f(f(x))=x$, and we get $f(xf(x)+y)=xf(x)+f(y)$.

$\textbf{Part 3.}$ With the same idea as above (to invoke inequalities), we see that $f(nxf(x)+y)=nxf(x)+f(y)$ for all $n\in \mathbb{Z}$, so $$t>nxf(x) \Rightarrow f(t)>nxf(x)$$Replace $t$ by $f(t)$, then $$f(t)>nxf(x) \Rightarrow t>nxf(x)$$This immediately implies $$nxf(x)<t\le (n+1)f(x) \iff nxf(x)<f(t)\le (n+1)xf(x)$$$$\Rightarrow |f(t)-t|\le xf(x)$$for all $t$ and $x$. It suffices to prove that $xf(x)$ is arbitrarily small. Suppose for some $C>0$, $xf(x)\ge C$ for all $x$, then $f(x)\ge\frac{C}{x}$. Then take $t=\frac{C}{n}, x=1$ for large integers $n>C$, so that $f(t)\ge n>1>t$, so $$n-\frac{C}{n}=\frac{C}{t}-t\le f(t)-t\le f(1)$$but take $n\rightarrow +\infty$ gives a contradiction.

So $xf(x)$ is arbitrarily small, implying $f(t)=t$ for all $t>0$.

QED
This post has been edited 8 times. Last edited by navi_09220114, Sep 9, 2024, 5:14 PM
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ItzsleepyXD
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#22
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By Lemma :
$\frac{f(f(x))}{x}$ is constant. So for some $c \in \mathbb{R^+} , f(f(x))=cx $ $\forall x \in \mathbb{R^+}$
$P(f(x),y): f(c^2xf(x)+y) = cxf(x)+f(y).$
so $c^3xf(x)+y=f(f(c^2xf(x)+y))=f(cxf(x)+f(y))=xf(x)+xy$.
implies that $c^3=1 \rightarrow c=1 \rightarrow f(f(x))=x$
$P(x,y) : f(xf(x)+y)=xf(x)+f(y) , P(x,f(y)) : f(xf(x)+f(y)) = xf(x)+y$

Claim $f(y)+xf(x) \geq y$
Proof : if $y>f(y)+xf(x)$ ,by $f(f(y))=y \rightarrow f(y) > y+xf(x)$
so $y-xf(x) > f(y) > y+xf(x)$ contradiction.

so $f(y)+xf(x) \geq y , y+xf(x) \geq f(y)$
implies that $y+xf(x) \geq f(y) \geq y-xf(x)$

Claim there is no $C \in \mathbb{R^+}$ such that $C \leq xf(x)$ $\forall x \in \mathbb{R^+}$
Proof : if there exist $C \leq xf(x)$ $\forall x \in \mathbb{R^+}$
from $y+xf(x) \geq f(y)$ implies that $y^2+yxf(x) \geq yf(y) \geq C$ but $y \rightarrow 0$ lead to a contradiction.
So $xf(x)$ is arbitrary small , thus $f(y) = y$ $\forall y \in \mathbb{R^+}$ . done $\square$
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jasperE3
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#23
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Functional_equation wrote:
Find all functions $f:\mathbb R^+\rightarrow \mathbb R^+$ such that$$f(f(x)f(f(x))+y)=xf(x)+f(y)$$for all $x,y\in \mathbb R^+$

$\textit{Proposed by Functional\_equation, Mr.C and TLP.39}$

Claim 1: $f(f(x))=cx$ for some constant $c\in\mathbb R^+$
True by the $fgh$ lemma (link).

Claim 2: $c=1$
Let $P(x,y)$ be the assertion $f(cxf(x)+y)=xf(x)+f(y)$. Note that $f(f(x))=cx$ implies:
$$f(cx)=f(f(f(x)))=cf(x)$$for all $x\in\mathbb R^+$. Then, suppose $c\ne1$, we have:
$P(x,cyf(y))\implies f(cxf(x)+cyf(y))=xf(x)+f(cyf(y))=xf(x)+cf(yf(y))$.
Swapping $x,y$ and comparing, we get that:
$$cf(xf(x))=xf(x)+d$$for some constant $d\in\mathbb R$. Then:
$$c^2d=c^3f(xf(x))-c^2xf(x)=cf(cxf(cx))-cxf(cx)=d,$$and so $d=0$ (as $c\ne1$).
The equation above turns into $cf(xf(x))=xf(x)$ now, plugging in $x=c$ we have $cf(cf(c))=cf(c)$, and by cancellation and injectivity $f(c)=1$.
Now spamming $f(f(x))=cx$ will give us our result: we have (applying $f$ to both sides) $f(1)=f(f(c))=c^2$, so $f\left(c^2\right)=f(f(1))=c$, so $f(c)=f\left(f\left(c^2\right)\right)=c^3$, so $c^3=1$, contradiction. Now that this claim has been proven, Claim 1 simplifies to $f(f(x))=x$.


Let $S=\{a\in\mathbb R^+\mid f(x+a)=f(x)+a\forall x\in\mathbb R^+\}$.
Claim 3: $\mathbb Q^+\subseteq S$
$P(1,1)\Rightarrow f(2f(1))=2f(1)$
From Claim 2 we have $xf(x)\in S$ for all $x\in\mathbb R^+$. In particular, $\frac1nf\left(\frac1n\right)\in S$ for all $n\in\mathbb N$ and $f(1)\in S$. Because of how $S$ is defined, if $a\in S$ then any natural multiple of $a$ must also be in $S$ (it's closed under addition), so $f\left(\frac1n\right)\in S$ and $2f(1)\in S$. Then, we have:
$$2f(1)+\frac1n=2f(1)+f\left(f\left(\frac1n\right)\right)=f\left(2f(1)+f\left(\frac1n\right)\right)=f(2f(1))+f\left(\frac1n\right)=2f(1)+f\left(\frac1n\right),$$so $f\left(\frac1n\right)=\frac1n$. Then $\frac1nf\left(\frac1n\right)\in S$ becomes $\frac1{n^2}\in S$, but remember all integral multiples of this figure are also in $S$, so $\frac{mn}{n^2}=\frac mn\in S$ for any $m,n\in\mathbb N$, hence proven.

Finish: $f(x)\ge x$
Suppose there is some $u\in\mathbb R^+$ with $f(u)<u$. Since $\mathbb Q^+$ is dense in $\mathbb R^+$, we can choose a rational $q$ such that $f(u)<q<u$. Recall that $q\in S$, so:
$$f(u)=f(u-q)+q>q,$$a contradiction.
Therefore $f(x)\ge x$ for all $x\in\mathbb R^+$. From $f(f(x))=x$ we obtain:
$$x=f(f(x))\ge f(x)\ge x$$with equality everywhere, hence $\boxed{f(x)=x}$ is the only solution (we can easily check that it fits).
This post has been edited 1 time. Last edited by jasperE3, Apr 28, 2025, 4:38 PM
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