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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
Local-global with Fibonacci numbers
MarkBcc168   26
N 4 minutes ago by pi271828
Source: ELMO 2020 P2
Define the Fibonacci numbers by $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n\geq 3$. Let $k$ be a positive integer. Suppose that for every positive integer $m$ there exists a positive integer $n$ such that $m \mid F_n-k$. Must $k$ be a Fibonacci number?

Proposed by Fedir Yudin.
26 replies
MarkBcc168
Jul 28, 2020
pi271828
4 minutes ago
Cauchy functional equations
syk0526   10
N 6 minutes ago by imagien_bad
Source: FKMO 2013 #2
Find all functions $ f : \mathbb{R}\to\mathbb{R}$ satisfying following conditions.
(a) $ f(x) \ge 0 $ for all $ x \in \mathbb{R} $.
(b) For $ a, b, c, d \in \mathbb{R} $ with $ ab + bc + cd = 0 $, equality $ f(a-b) + f(c-d) = f(a) + f(b+c) + f(d) $ holds.
10 replies
syk0526
Mar 24, 2013
imagien_bad
6 minutes ago
Three circles are concurrent
Twoisaprime   21
N 16 minutes ago by L13832
Source: RMM 2025 P5
Let triangle $ABC$ be an acute triangle with $AB<AC$ and let $H$ and $O$ be its orthocenter and circumcenter, respectively. Let $\Gamma$ be the circle $BOC$. The line $AO$ and the circle of radius $AO$ centered at $A$ cross $\Gamma$ at $A’$ and $F$, respectively. Prove that $\Gamma$ , the circle on diameter $AA’$ and circle $AFH$ are concurrent.
Proposed by Romania, Radu-Andrew Lecoiu
21 replies
+1 w
Twoisaprime
Feb 13, 2025
L13832
16 minutes ago
IMO Shortlist 2011, Algebra 3
orl   45
N 23 minutes ago by Ilikeminecraft
Source: IMO Shortlist 2011, Algebra 3
Determine all pairs $(f,g)$ of functions from the set of real numbers to itself that satisfy \[g(f(x+y)) = f(x) + (2x + y)g(y)\] for all real numbers $x$ and $y$.

Proposed by Japan
45 replies
1 viewing
orl
Jul 11, 2012
Ilikeminecraft
23 minutes ago
Hard FE with positive reals
egxa   8
N 33 minutes ago by megarnie
Source: Turkey Olympic Revenge 2023 Shortlist A4
Find all functions $f:\mathbb{R^+}\to \mathbb{R^+}$ such that for all $x,y\in \mathbb{R^+}$
$f(xf(y)+y)=f(f(y))+yf(x)$
Proposed by Şevket Onur Yılmaz
8 replies
egxa
Jan 22, 2024
megarnie
33 minutes ago
Like Father Like Son... (or Like Grandson?)
AlperenINAN   1
N 37 minutes ago by hakN
Source: Turkey TST 2025 P4
Let $a,b,c$ be given pairwise coprime positive integers where $a>bc$. Let $m<n$ be positive integers. We call $m$ to be a grandson of $n$ if and only if, for all possible piles of stones whose total mass adds up to $n$ and consist of stones with masses $a,b,c$, it's possible to take some of the stones out from this pile in a way that in the end, we can obtain a new pile of stones with total mass of $m$. Find the greatest possible number that doesn't have any grandsons.
1 reply
AlperenINAN
Today at 6:09 AM
hakN
37 minutes ago
Crazy number theory
MTA_2024   5
N 38 minutes ago by bjump
Find all couple $(p;q)$ of primes (greater than 5) such that : $$pq \mid (5^q-3^q)(5^p-3^p)$$
5 replies
MTA_2024
3 hours ago
bjump
38 minutes ago
hard number theory problem
Zavyk09   0
an hour ago
Source: forgotten
Find all couple $(x, y)$ of positive integers such that:
$$2^n + 3^n \mid x^n + y^n, \forall n \in \mathbb{N}^*$$
0 replies
Zavyk09
an hour ago
0 replies
The return of a legend inequality
giangtruong13   2
N an hour ago by polishedhardwoodtable
Source: Legacy
Given that $0<a,b,c,d<1$.Prove that: $$ (1-a)(1-b)(1-c)(1-d) > 1-a-b-c-d $$
2 replies
giangtruong13
2 hours ago
polishedhardwoodtable
an hour ago
Slightly weird points which are not so weird
Pranav1056   9
N an hour ago by Retemoeg
Source: India TST 2023 Day 4 P1
Suppose an acute scalene triangle $ABC$ has incentre $I$ and incircle touching $BC$ at $D$. Let $Z$ be the antipode of $A$ in the circumcircle of $ABC$. Point $L$ is chosen on the internal angle bisector of $\angle BZC$ such that $AL = LI$. Let $M$ be the midpoint of arc $BZC$, and let $V$ be the midpoint of $ID$. Prove that $\angle IML = \angle DVM$
9 replies
Pranav1056
Jul 9, 2023
Retemoeg
an hour ago
2023 factors and perfect cube
proxima1681   4
N an hour ago by anudeep
Source: Indian Statistical Institute (ISI) UGB 2023 P4
Let $n_1, n_2, \cdots , n_{51}$ be distinct natural numbers each of which has exactly $2023$ positive integer factors. For instance, $2^{2022}$ has exactly $2023$ positive integer factors $1,2, 2^{2}, 2^{3}, \cdots 2^{2021}, 2^{2022}$. Assume that no prime larger than $11$ divides any of the $n_{i}$'s. Show that there must be some perfect cube among the $n_{i}$'s.
4 replies
proxima1681
May 14, 2023
anudeep
an hour ago
circle geometry showing perpendicularity
Kyj9981   1
N an hour ago by Retemoeg
Two circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$. A line through $B$ intersects $\omega_1$ and $\omega_2$ at points $C$ and $D$, respectively. Line $AD$ intersects $\omega_1$ at point $E \neq A$, and line $AC$ intersects $\omega_2$ at point $F \neq A$. If $O$ is the circumcenter of $\triangle AEF$, prove that $OB \perp CD$.
1 reply
Kyj9981
Today at 11:53 AM
Retemoeg
an hour ago
Hard problem
Tendo_Jakarta   5
N an hour ago by Tendo_Jakarta
Let the sequence \(x_{n}\) be such that
\[u_{1} = 1; \quad u_{n+1} = \dfrac{u_{1} + u_{2} +...+u_{n}}{n}+n-1 \quad \forall n \in \mathbb{N^{*}}\]and \(y_{n} =\dfrac{1}{u_{1}u_{2}} + \dfrac{1}{u_{3}u_{4}} + ... + \dfrac{1}{u_{2n-1}u_{2n}}  \quad \forall n \geq 1\). Find \(\lim_{n\rightarrow\infty}{y_{n}}\).
5 replies
Tendo_Jakarta
2 hours ago
Tendo_Jakarta
an hour ago
Oh no! Inequality again?
mathisreaI   108
N an hour ago by Maximilian113
Source: IMO 2022 Problem 2
Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$, there is exactly one $y \in \mathbb{R}^+$ satisfying $$xf(y)+yf(x) \leq 2$$
108 replies
mathisreaI
Jul 13, 2022
Maximilian113
an hour ago
interesting function
menpo   15
N Dec 15, 2024 by megarnie
Source: 9th EMC, 12th December 2020 - 20th December 2020, SENIOR league, P4.
Let $\mathbb{R^+}$ denote the set of all positive real numbers. Find all functions $f: \mathbb{R^+}\rightarrow \mathbb{R^+}$ such that
$$xf(x + y) + f(xf(y) + 1) = f(xf(x))$$for all $x, y \in\mathbb{R^+}.$

Proposed by Amadej Kristjan Kocbek, Jakob Jurij Snoj
15 replies
menpo
Dec 22, 2020
megarnie
Dec 15, 2024
interesting function
G H J
G H BBookmark kLocked kLocked NReply
Source: 9th EMC, 12th December 2020 - 20th December 2020, SENIOR league, P4.
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menpo
209 posts
#1 • 10 Y
Y by Functional_equation, centslordm, HWenslawski, chessgocube, MathLuis, PRMOisTheHardestExam, megarnie, kikimyrt, Lasier, cubres
Let $\mathbb{R^+}$ denote the set of all positive real numbers. Find all functions $f: \mathbb{R^+}\rightarrow \mathbb{R^+}$ such that
$$xf(x + y) + f(xf(y) + 1) = f(xf(x))$$for all $x, y \in\mathbb{R^+}.$

Proposed by Amadej Kristjan Kocbek, Jakob Jurij Snoj
This post has been edited 2 times. Last edited by menpo, Dec 22, 2020, 12:50 PM
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IndoMathXdZ
691 posts
#2 • 11 Y
Y by Functional_equation, square_root_of_3, RevolveWithMe101, GorgonMathDota, Morskow, mijail, centslordm, Kobayashi, chessgocube, PRMOisTheHardestExam, cubres
One of the hardest FE problems I've done. Got this after a lot of hours after the test and some hints :(
The answer is $f(x) = \frac{1}{x}$ which satisfies since
\[ xf(x+y) + f(xf(y) + 1) = \frac{x}{x+y} + \frac{y}{x+y} = 1 = f(xf(x)) \]for all $x,y \in \mathbb{R}^+$.
We now prove that there are no other functions that satisfy the functional equation.
Let $P(x,y)$ be the assertion of $x$ and $y$ to the given functional equation.
Claim 01. $f(x) < f(f(1))$ for all $x > 1$.
Proof. $P(1,y)$ gives us $f(y+1) + f(f(y) + 1) = f(f(1)) \Rightarrow f(y+1) < f(f(1)) \ \forall y \in \mathbb{R}^+$, which proves our claim.
Claim 02. $f$ is injective.
Proof. Suppose otherwise, that there exists $a < b$ such that $f(a) = f(b)$. $P(x,a)$ and $P(x,b)$ gives us
\[ f(x+a) = f(x+b) \ \forall x \in \mathbb{R}^+ \]Thus, $f$ is periodic for all $x \ge c$, a constant $c$ with period $b - a$.
Fix $x_1, y_1 \in \mathbb{R}^+$, $x_1 > c$ and pick $n \in \mathbb{N}$ such that $(x_1 + np)f(x_1 + y_1) \ge f(f(1))$ and $(x_1 + np)f(x_1) > 1$.
Thus, $P(x_1 + np, y)$ gives us
\[ (x_1 + np)f(x_1 + y_1) + f((x_1 + np)f(y_1) + 1) = f((x_1 + np)f(x_1)) \]However, this forces $f((x_1 + np)f(x_1)) > (x_1 + np) f(x_1 + y_1) \ge f(f(1))$, and since $(x_1 + np)f(x_1) > 1$, from the previous claim, we get that
\[ f((x_1 + np)f(x_1)) \le f(f(1)) \]a contradiction.
Claim 03. $f$ is involutive.
Proof. $P(1,f(y))$ and $P(1,y)$ gives us
\[ f(f(y) + 1) + f(f(f(y)) + 1) = f(y + 1) + f(f(y) + 1) \Rightarrow f(y + 1) = f(f(f(y)) + 1) \Rightarrow f(f(y)) = y \ \forall y \in \mathbb{R}^+ \]
Claim 04. $xf(x) \le 1$.
Proof. If there exists $a \in \mathbb{R}^+$ such that $af(a) > 1$, then $P(x,f(y))$ gives us
\[ xf(x + f(y)) + f(xy + 1) = f(xf(x)) \]Use the cancelling trick on $P(a,y)$ and let $ay + 1 = af(a)$, which is possible as $af(a) > 1$, we get a contradiction.
Claim 05. $f(xf(x)) = 1$ for all $x \in \mathbb{R}^+$.
Proof. First, we prove $f(xf(x)) \le 1$ for all $x \in \mathbb{R}^+$.
Assume otherwise, that there exists $b \in \mathbb{R}^+$ such that $f(bf(b)) > 1$. Notice that $P(b,y)$ gives us
\[ f(bf(b)) = bf(b+y) + f(bf(y) + 1) \overset{ f(bf(y) + 1) < f(f(1)) = 1}{\le} bf(b+y) + 1 \overset{f(b+y) \le \frac{1}{b+y}}{\le} \frac{b}{b+y} + 1 \]Now, we have $f(bf(b)) \le \lim_{y \to \infty} \frac{b}{b + y} + 1$, which forces $f(bf(b)) \le 1$, a contradiction.
Now, assume that there exists $c \in \mathbb{R}^+$ such that $f(cf(c)) < 1$. Let $f(cf(c)) = 1 - \varepsilon$. Take a constant $\Delta$ such that for all $y \ge \Delta$, then $f(y+1) < \varepsilon$. Let $\delta > \Delta$. From $P(1,\delta)$, we have
\[ f(f(\delta) + 1) > 1 - \varepsilon \]Remembering $f(x,f(y))$ gives us $f(xf(x)) = f(xy + 1) + xf(x+f(y)) >  f(xy + 1)$, then this motivates $P \left( c, f \left( \frac{f(\delta)}{c} \right) \right)$ which gives us
\[ 1 - \varepsilon = f(cf(c)) > f(f(\delta) + 1) > 1 - \varepsilon \]a contradiction.
Therefore $f(xf(x))$ is a constant. Since $f$ is injective, then $xf(x)$ is a constant as well, which means $f(x) = \frac{c}{x}$ is the only possible solution. Checking back,
\[ 1 = f(xf(x)) = xf(x+y) + f(xf(y) + 1) = \frac{cx}{x + y} + f \left( \frac{cx}{y} + 1 \right) = \frac{cx}{x + y} + \frac{cy}{cx + y} \]for all $x,y \in \mathbb{R}^+$.
Take $x = y = 1$, gives us $\frac{c}{2} + \frac{c}{c + 1} = 1$, which means $c(c + 1) = 2$. This gives $c = -2$ or $c = 1$. But since $c \in \mathbb{R}^+$, this gives $c = 1$.
Hence, the only solution is $\boxed{f(x) = \frac{1}{x}}$, done.
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mela_20-15
125 posts
#7 • 4 Y
Y by centslordm, chessgocube, PRMOisTheHardestExam, cubres
As proven above $f$ is involutive and $xf(x)\le 1$ . Set $c=f(1)$ :$f(c)=1$ we have $c\le 1$.


We will need the following get the four equations from plugging in $(x,y),(y,x),(f(x),f(y)),(f(y),f(x))$ and compare them to get :
$$(x+y)f(x+y)=(f(x)+f(y))f(f(x)+f(y)).....(2)$$set $(c,f(y))$ and get $$1=cf(c+f(y))+f(cy+1)\le cf(c+f(y))+\frac{1}{cy+1}$$$$f(c+f(y))\ge \frac{1}{c+1/y}$$and use $(2)$ to get that
$(y+1)f(y+1)=(c+f(y))f(c+f(y))\ge \frac{c+f(y)}{c+1/y}$ in other words $\lim_{y\to \infty}yf(y)=1$.
Next use relation (from setting $x=1$) $ f(y+1)+f(f(y)+1)=1 $ and plug these into $(2)$ to get that $$(y+f(y)+2)f(y+f(y)+2)=c$$use $(2)$ again to get that $$(f(y+2)+y)f(f(y+2)+y)=(y+2+f(y))f(y+2+f(y))=c$$I rewrite the latter two relations $$(f(y+2)+y+4)f(f(y+2)+y+4)=(f(y+2)+y)f(f(y+2)+y)=c$$Now plug in the equation $(f(y+2)+y,4)$ to finally get that $$\frac{c(f(y+2)+y)}{f(y+2)+y+4}+f((f(y+2)+y)f(4)+1)=1$$It is easy to see that if we let $y\to \infty$ we will get that $c=1$ (we could have done something similar in other ways too).
After that set $y=1$ get that $f(xf(x))=(x+1)f(x+1)$ then $f$-this to get $xf(x)=(x+2)f(x+2)$ and $(x+2m)f(x+2m)=xf(x)$ for any integer $m$. This and the fact that $\lim_{y\to \infty}yf(y)=1$ give that $xf(x)=1$ for any $x$.
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Supercali
1260 posts
#8 • 5 Y
Y by W.R.O.N.G, centslordm, chessgocube, PRMOisTheHardestExam, cubres
Let $P(x,y)$ denote the given proposition.

Claim 1: $f(x)>1$ $\implies$ $x<1$.
Proof: Assume the contrary, i.e., $f(x)>1$ and $x \geq 1$ for some $x$. Let $y=x(f(x)-1)>0$. Then $P(x,y)$ gives $$f(xf(x)) > xf(x+y) \geq f(xf(x))$$Contradiction! $\blacksquare$

In particular, by Claim 1, if $f(x)>x$ for some $x$, then we must have $x <1$; otherwise, we would have $f(x) >x \geq 1$, which contradicts Claim 1.

Claim 2: $f$ is injective.
Proof: Assume FTSOC that there exist $a>b$ such that $f(a)=f(b)$. Then comparing $P(x,a)$ and $P(x,b)$, we get $f(x+a)=f(x+b)$ for all $x$ $\implies$ $f$ is eventually periodic with period $p=a-b>0$, Now, $P(x,p)$ gives $$f(xf(x))>xf(x+p)=xf(x)$$$\implies$ $xf(x)<1$ for all $x$ by Claim 1. But then, $$f(x)=f(x+np) < \frac{1}{x+np}$$for all positive integers $n$, and we obtain a contradiction by taking $n \rightarrow +\infty$. $\blacksquare$

Claim 3: $f(f(y)) =y$ for all $y$.
Proof: $P(1,y)$ gives $$f(y+1)+f(f(y)+1)=f(f(1))$$for all $y$. Putting $f(y)$ instead of $y$ in the above and using injectivity, we get $f(f(y))=y$. $\blacksquare$

Claim 3 in particular implies that $f$ is bijective.

Claim 4: $f(x) \leq \frac{1}{x}$ for all $x$.
Proof: Assume $xf(x)>1$ for some $x$. Then by surjectivity we can choose a $y$ such that $xf(y)+1=xf(x)$. But now $P(x,y)$ gives $f(xf(x))>f(xf(x))$, contradiction! $\blacksquare$

Let $S=\{f(x) \mid x>1 \}$. Let $m= \sup S$, which exists since $f(x)<1$ for all $x>1$ by Claim 1 and Claim 3. Note that $z\in S$ $\implies$ $f(z)>1$ by Claim 3.

Claim 5: $f(xf(x)) \geq m$ for all $x$.
Proof: For any $x>0$, and for any $z \in S$, there exists a $y>0$ such that $xf(y)+1=f(z)$ by surjectivity. Then $P(x,y)$ $\implies$ $f(xf(x)) > f(f(z))=z$. Since this holds for all $z \in S$, we must have $f(xf(x)) \geq m$ for all $x$. $\blacksquare$

Claim 6: $f(xf(x)) \leq m$ for all $x$.
Proof: Fix an $x$. Then $P(x,y)$ gives $$f(xf(x))=xf(x+y)+f(xf(y)+1) \leq \frac{x}{x+y} +m$$for all $y$. Then taking $y \rightarrow \infty$ in the above, we get $f(xf(x)) \leq m$. $\blacksquare$

Claim 5 and Claim 6 give $f(xf(x))=m$ for all $x$. Therefore injectivity gives $f(x)=\frac{c}{x}$ for some constant $c$. Putting this in $P(x,y)$, we get that $c=1$. Therefore $$\boxed{f(x)=\frac{1}{x} \ \ \forall x \in \mathbb{R}^+}$$
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Aryan-23
558 posts
#9 • 7 Y
Y by 606234, centslordm, mijail, chessgocube, PRMOisTheHardestExam, Pranav1056, cubres
Solution
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MathLuis
1450 posts
#10 • 5 Y
Y by chessgocube, rama1728, PRMOisTheHardestExam, CyclicISLscelesTrapezoid, cubres
AWSOME F.E.!! (my respect to the creators of this beauty)
Solved with Rama1728 :pilot: (Caling $P(x,y)$ assertion and blah blah blah)
Claim 1: if $f(x)>1$ then $x<1$
Proof: Use $P(x,x(f(x)-1))$ and assume that $f(x)>1$, now if $x>1$ then
$$f(xf(x(f(x)-1))+1)=(1-x)f(xf(x))<0 \text{contradiction!!}$$For proving the next claim we might need $f(x)>1$ then $x<1$ which means $f(x)>x$ then $x<1$.
Claim 2: $f$ is injective.
Proof: Assume that its not injective then there exists $a>b$ such that $f(a)=f(b)$ and by $P(x,a)-P(x,b)$ for $x>b$
$$f(x+a)=f(x+b) \implies f \; \text{periodic with period} \; a-b$$Now using $P(x,a-b)$ and Claim 1 we get
$$f(xf(x))=xf(x)+f(xf(a-b)+1)>xf(x) \implies f(x)<\frac{1}{x} \implies f(x)<\frac{1}{x+(a-b)k} \implies f(x)<0 \; \text{contradiction!!}$$On the last step $k$ was a posititve integer and we just let $k \to \infty$ to get the contradiction.
Claim 3: $f$ is an involution.
Proof: Use $P(f(y),1)-P(y,1)$
$$f(y+1)+f(f(y)+1)=f(f(y)+1)+f(f(f(y))+1) \implies f(y+1)=f(f(f(y))+1) \implies f(f(y))=y \implies f \; \text{involution}$$Claim 4: $f(x) \le \frac{1}{x}$
Proof: Just use $P \left(x, f \left(\frac{xf(x)-1}{x} \right) \right)$ and assume that there exists $x$ such that $f(x)>\frac{1}{x}$
$$xf \left(x+f \left(\frac{xf(x)-1}{x} \right) \right)=0 \; \text{contradiction!!}$$Claim 5: $f(n)=\frac{1}{n}$ for every posititve integer $n$.
Proof: Using Claim 4 we get $f(1) \le 1$ and also assume that there exists $x$ such that if $x<1$ then $f(x)<1$. and this will mean that if $f(x)<1$ then $x<1$ but note that using involution on Claim 1 this will mean that $1>x>1$ which its not possible. Hence if $x<1$ then $f(x)>1$ and using Claim 3 on this we get if $f(x)<1$ then $x>1$ so this with Claim 1 and Claim 3 forces that $f(1)=1$. Now by $P(n,1)$ and easy induction.
$$(n+1)f(n+1)=f(nf(n)) \implies f(n)=\frac{1}{n}$$On the final part we will use this when $n \to \infty$ so this means $\lim_{x \to \infty} f(x)=0$
Final part: Using Claim 4 on the F.E. and letting $y \to \infty$ we get
$$f(xf(x)) \le \lim_{y \to \infty} \frac{x}{x+y}+\frac{1}{xf(y)+1}=1$$But using Claim 4 anda variant of Claim 1 we get that $f(xf(x)) \ge 1$.
Hence the unique solution of the F.E. is $f(x)=\frac{1}{x}$ thus we are done :blush:
This post has been edited 2 times. Last edited by MathLuis, Dec 10, 2021, 5:23 PM
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IvoBucata
46 posts
#11 • 1 Y
Y by cubres
MathLuis wrote:
Proof: Assume that its not injective then there exists $a>b$ such that $f(a)=f(b)$ and by $P(x,a)-P(x,b)$
$$f(x+a)=f(x+b) \implies f \; \text{periodic with period} \; a-b$$Now using $P(x,a-b)$ and Claim 1 we get
$$f(xf(x))=xf(x)+f(xf(a-b)+1)>xf(x) \implies f(x)<\frac{1}{x} \implies f(x)<\frac{1}{x+(a-b)k} \implies f(x)<0 \; \text{contradiction!!}$$

Excuse me if I am mistaken, but how exactly does $P(x,a-b)$ imply $f(xf(x))=xf(x)+f(xf(a-b)+1)$ for $x<b$?
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MathLuis
1450 posts
#12 • 2 Y
Y by rama1728, cubres
Oops, i forgot to mention that $x>b$ (Sorry)
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CyclicISLscelesTrapezoid
371 posts
#13 • 5 Y
Y by RP3.1415, Bradygho, rama1728, Pranav1056, cubres
I think this solution is new.

Let $P(x,y)$ be the assertion that \[xf(x+y)+f(xf(y)+1)=f(xf(x))\]
Claim 1: If positive real numbers $x$ and $a$ satisfy $x>a$, then $f(x)<\frac{f(af(a))}{a}$.
Proof: By $P(a,x-a)$, we have \[f(af(a)=af(x)+f(af(x-a)+1)>af(x),\]so $f(x)<\frac{f(af(a))}{a}$.

If $f(a)>1$, then we can let $x=af(a)$ in the above claim to get $a<1$. Taking the contrapositive of this statement, we see that $f(a) \leq 1$ if $a \geq 1$.

Claim 2: If positive real numbers $x$ and $a$ satisfy $x>a$, then $f(x) \geq \frac{f(af(a))-1}{a}$
Proof: Notice that since $xf(y)+1 \geq 1$, we have $f(xf(y)+1) \leq 1$. By $P(a,x-a)$, we have \[af(x)=f(af(a))-f(af(x-a)+1) \geq f(af(a))-1,\]so $f(x) \geq \frac{f(af(a))-1}{a}$

Claim 3: $\lim_{x \to \infty}f(x)=0$.
Proof: We first show that $\lim_{x \to \infty}f(x)$ exists. Consider the intervals \[\mathcal{A}_n=\left[\frac{f(nf(n))-1}{n},\frac{f(nf(n))}{n}\right)\]for positive integers $n$. We know that $f(x)$ lies in $\mathcal{A}_n$ if $x>n$. We claim that there exists a real number $L$ in $\mathcal{A}_n$ for every positive integer $n$. Assume FTSOC that there doesn't exist such $L$. Then, there must be a positive integer $m$ such that $\mathcal{A}_1 \cap \mathcal{A}_2 \cap \cdots \cap \mathcal{A}_m$ is nonempty, but $\mathcal{A}_1 \cap \mathcal{A}_2 \cap \cdots \cap \mathcal{A}_{m+1}$ is empty. However, that means $f(x)$ cannot exist for any $x>m+1$, a contradiction. Thus, there exists such $L$. Now, notice that for any real number $x \geq 2$, $L$ and $f(x)$ both lie in the interval $\mathcal{A}_{\lfloor x \rfloor-1}$, so $|L-f(x)|<\frac{1}{\lfloor x \rfloor-1}$. Since this converges to $0$, $\lim_{x \to \infty}f(x)$ must converge to $L$.

Assume FTSOC that $L \neq 0$. By $P(x,y)$ for arbitrarily large $x$ and $y$, we obtain $xL+f(Lx+1)=f(Lx)$, a contradiction since the LHS grows unbounded while the RHS converges to $L$. Thus, $L=0$.

Claim 4: $f$ is injective.
Proof: Assume FTSOC that there exists positive real numbers $a$ and $b$ such that $f(a)=f(b)$ and $a \neq b$. We compare $P(x,a)$ and $P(x,b)$ to get $f(x+a)=f(x+b)$, so $f$ is eventually periodic. However, that means $\lim_{x \to \infty}f(x)$ cannot converge to $0$, a contradiction.

Now, take $P(a,y)$ for fixed $a$ and arbitrarily large $y$. Then, we have \[\lim_{x \to 1}f(x)=f(af(a)),\]so $f(af(a))$ is constant. Thus, $f(xf(x))=f(yf(y))$ for all real numbers $x$ and $y$, so $xf(x)=yf(y)$. Therefore, we have $f(x) \equiv \frac{c}{x}$, and it's easy to obtain that $f(x) \equiv \frac{1}{x}$ is the only solution.
This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Dec 18, 2022, 7:40 PM
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ZETA_in_olympiad
2211 posts
#14 • 1 Y
Y by cubres
This was hard. Denote the assertion by $P(x,y).$ We solve the problem in a number of steps.

Step 1: We show that $f(x)>1 \implies x<1.$
Assume not. Then $P(x,xf(x)-x)$ readily yields contradiction. As a corollary note that $f(x)>x$ if $x>1.$

Step 2: We show that $f$ is one to one.
If $f(u)=f(v).$ Then $P(x,u)$ and $P(x,v)$ imply $f$ is periodic with period $u-v.$ Now $P(x,u-v)$ gives $f(xf(x))>xf(x).$ This means $f(x)<1/(x+c(u-v))$ if $u-v\neq 0.$ Take large $c$ to get contradiction.

Step 3: We show that $f$ is an involution.
This is obtained when $P(x,f(y))$ is compared with $P(x,y).$

Step 4: We show that $xf(x)\leq 1.$
If not then take such a $z.$ Then $P(z,f((zf(z)-1)/z))$ readily yields contradiction.

Step 5: We show that $f(xf(x))\leq 1.$
We have $f(xf(x))=f(xf(y)+1)+xf(x+y)<x/(x+y)+1.$ Taking large $y$ forces $f(xf(x))\leq 1.$

Step 6: We show that $f(xf(x))=1.$
Assume that we have some positive $k$ and $z$ such that $f(zf(z))=1-k.$ If $v$ is bounded then $P(1,v)$ implies $f(f(v)+1)=1-f(v+1)>1-k.$ We have contradiction from $P(z,f(f(v)/z)).$

Step 7: We show $f(x)=1/x.$
We have that $f(x)=c/x$ for some constant $c.$ Checking gives $c=1$ and thus $f(x)=1/x$ which can be seen to work.
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L567
1184 posts
#15 • 4 Y
Y by Mango247, Mango247, Mango247, cubres
Solved with Aayuu and TheProblemIsSolved

As usual, let $P(x,y)$ denote the given assertion. Suppose $f(x) > 1$ for some $x$, then $xf(x) > x$ and so $P(x,xf(x)-x)$ gives that $(1-x)f(xf(x)) = f(xf(y)+1) > 0$ so $x < 1$.

Suppose $f(a) = f(b)$ for distinct $a,b$ then comparing $P(x,a), P(x,b)$ we have that $f(x+a) = f(x+b)$ for all $x$, which means $f$ is eventually periodic with period $p = |a-b|$. Taking $P(x,p)$ gives that $f(xf(x)) > xf(x)$. If $xf(x) > 1$, then by the previous paragraph, we get a contradiction, so actually $f(x) \leqslant \frac{1}{x}$ for all big enough $x$. But $f(x) = f(x+kp)$ so $f(x) \leqslant \frac{1}{x+kp}$ for all $k$, impossible as $f$ is positive, so $f$ must be injective.

Now, compare $P(1,y)$ and $P(1,f(y))$ to get that $f(f(y)) = y$ for all $y$, due to injectivity. Note that since $f$ is surjective, if $f(x) > \frac{1}{x}$, then we can choose $f(y)$ so that $xf(y) + 1 = xf(x)$, giving a contradiction. Therefore, we have that $f(x) \leqslant \frac{1}{x}$ for all $x$.

Claim: $\lim_{y \rightarrow \infty} f(xf(y)+1)$ exists and equals $f(xf(x))$

Proof: Note that $$f(xf(x)) - \frac{x}{x+y} \leqslant f(xf(y)+1) < f(xf(x))$$using the fact that $0 < xf(x+y) \leqslant \frac{x}{x+y}$. So taking $y$ large enough makes the claim true. $\square$

Since $\lim_{y \rightarrow \infty} f(y) = 0$ $\left(0 < f(y) \leqslant \frac{1}{y} \right)$, we must have, by the above claim, that $f(xf(x)) = \lim_{y \rightarrow 1^{+}} f(y)$, which is constant. So $xf(x)$ must be constant, and checking gives that only $f(x) = \frac{1}{x}$ works, so done. $\blacksquare$
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YaoAOPS
1486 posts
#16 • 1 Y
Y by cubres
Denote the assertion with $P(x, y)$.
Then, by $P(1, x)$ \[ f(1 + x) + f(f(x) + 1) = f(f(1)) \]so $1 + x \ne f(1)$ over all $x$ and thus $f(1) \le 1$.

Claim: $f(x) \le \frac{c}{x}$ for sufficiently large $x$ and some $c$.
Proof. We have that \[ f(x + y) < \frac{f(xf(x))}{x} \]so $f(f(1))$ bounds $f$ on $(1, \infty)$.
As such, this means that \[ f(x + y) > \frac{f(xf(x)) - f(f(1))}{x}. \]Let $c = \max\{1, f(f(1))\}$.
For each $x$, if $f(x) < \frac{1}{x}$ then this follows. Else, if $f(x) > \frac{1}{x}$ then \[ f(x) < \frac{f(xf(x))}{x} < \frac{f(f(1))}{x} \]$\blacksquare$
As such, \[ \lim_{x \to \infty} f(x) = 0. \]
Claim: $f$ is injective.
Proof. Suppose $f(a) = f(b)$ for $a \ne b$.
Then, by $P(x, a)$ and $P(x, b)$ it follows that \[ f(x + a) = f(x + b) \]and thus $f$ is periodic. However, this contradicts the limit. $\blacksquare$
Taking $y \to \infty$ gives that
\[
	\lim_{y \to 0} f(1 + y) = f(xf(x))
\]where the limit exists since $f(xf(y) + 1) \le f(xf(x))$ holds.
This is independent from $x$, so $xf(x)$ is constant and thus $f(x) = \frac{a}{x}$, of which only $a = 1$ works.
This post has been edited 2 times. Last edited by YaoAOPS, Jun 5, 2023, 4:59 PM
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DS68
29 posts
#17 • 1 Y
Y by cubres
Let $P(x,y)$ be the assertion above.
First thing is we know that for all $x > 1$ we have $f(x) \leq 1$ as by taking $x+y = xf(x)$, it shows us if $f(x) > 1$, we must have $x < 1$ to avoid a contradiction in $f(xf(y)+1) = (1-x)f(xf(x)) > 0$. Hence if $f(x) > 1, x > 1$, this would be a contradiction.

Now let us assume $f$ is not injective. Then we get $f(x+a) = f(x+b)$ for some certain $a,b$. We must have at some point $f(x) \leq \frac{1}{x}$ for all $x \geq B$. Notice, if this is not true, we have $f(c) > \frac{1}{c}$ for some arbitrarily large $c$. Then $cf(c) > 1$, which means
\[cf(c+y) + f(cf(y) + 1) < f(cf(c)) < 1\]and $f(c+y) < \frac{1}{c}$. But this is a contradiction as we have by the not-injective condition, a periodicity above some lower bound where we can assume $c$ is some arbitrarily large number, giving $f(c+p) = f(c) > \frac{1}{c}$.

Thus now, notice even if $xf(x) \leq 1$ for every $x \geq B$, we have by the periodicity, taking a $x+Np$, we get that $(x+Np)f(x+Np) = (x+Np)f(x)$, which must increase arbitrarily as $N$ increases, thus breaking the upper bound of $1$. This must mean we have $f$ is injective.

$P(1,y)$,
\[f(y+1) + f(f(y)+1) = f(f(1))\]$y \mapsto f(y)$,
\[f(f(y)+1) + f(f(f(y))+1) = f(f(1))\]
\[f(y+1) = f(f(f(y))+1)\]\[y = f(f(y))\]which gives bijective and involutive.
Now then since we have involutive, we know all values of larger than $1$ is reached by numbers smaller than $1$ and vice versa.
Notice by forcing $xf(y)+1 = xf(x)$, we know $f(y) = f(x) - \frac{1}{x}$ can not be made, giving $f(x) \leq \frac{1}{x}$ for all $x$. Hence $\lim_{x \rightarrow \infty} f(x) = 0$. Hence taking $y \rightarrow \infty$, we get $C = f(xf(x))$, where we know a subsequential limit exists as $f(xf(y)+1)$ is bounded. Hence $xf(x) = c \rightarrow f(x) = \frac{c}{x}$, subbing to get $f \equiv \frac{1}{x}$.
This post has been edited 4 times. Last edited by DS68, Dec 21, 2023, 11:29 AM
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IAmTheHazard
5000 posts
#18 • 2 Y
Y by tadpoleloop, cubres
I think this is different.


The answer is $f(x)=1/x$ only, which works since both sides equal $1$. Let $P(x,y)$ denote the assertion. From $P(1,y)$, we have $f(y+1)+f(f(y)+1)=f(f(1))$, hence $f(y+1)<f(f(1))$ for all $y$, or equivalently $f(y)<f(f(1))$ for $y>1$.

Claim: $f$ is injective.
Proof: Suppose $f(a)=f(b)$ with $a \neq b$. By comparing $P(x,a)$ with $P(x,b)$ we find that $f(x+a)=f(x+b)$, hence $f$ is eventually periodic with period $|a-b|:=T$. Now by considering $P(kT,1)$ for some integer $k$ and sending $k \to \infty$, $kTf(kT+1)$ goes to infinity as well since $f(kT+1)$ is eventually constant. Thus $f(kTf(kT))$ is unbounded as $k \to \infty$, but $f(kT)$ is eventually constant as well, hence we will certainly have $kTf(kT)>1$ eventually. But this contradicts $f(y)<f(f(1))$. $\blacksquare$

Revisiting $P(1,y)$ and comparing it with $P(1,f(y))$ yields $f(y+1)=f(f(f(y))+1)$, hence by injectivity $f(f(y))=y$, i.e. $f$ is an involution. Thus we can rewrite the assertion (by replacing $y$ with $f(y)$) as $xf(x+f(y))+f(xy+1)=f(xf(x))$. This implies that $xf(x) \leq 1$ for all $x$, since otherwise we may pick $y$ with $xy+1=xf(x)$ and get a contradiction.

Now suppose that we have $a,b$ with $f(af(a))>f(bf(b))$. By comparing $P(a,y)$ with $P(b,ay/b)$ and subtracting, it follows that $af(a+f(y))-b(b+f(ay/b))=f(af(a))-f(bf(b))>0$. On the other hand, since $f$ is an involution we can make $f(y)$ arbitrarily large, and since $xf(x) \leq 1$ for all $x$, it follows that $af(a+f(y))$ becomes arbitrarily small. But it must always be at least $f(af(a))-f(bf(b))$, which is a contradiction. Hence $f(xf(x))$ is constant, so by injectivity $xf(x)$ is as well and $f(x)=c/x$ for some $c$. It is straightforward to plug in and verify that only $c=1$ works. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Oct 7, 2023, 10:17 PM
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Marius_Avion_De_Vanatoare
54 posts
#19 • 1 Y
Y by cubres
One of the most known $\mathbb{R}_+$ FE.
As usual, let $P(x;y)$ be the assertion. Let $f(f(1))=k$.
$P(1;y) \Rightarrow f(y+1)+f(f(y)+1)=k$, in particular that $f(x)<k, \forall x>1$.
First, I will prove that $f(xf(x)) \le k$. Assume it is not the case and we have some $x_1$, which does not fit. $P(x_1;y) $ Together with $f(x_1f(y)+1) < k$ results in $f$ being bounded below by something bigger than $0$, for large enough inputs. Now putting back large $x$ and $y$ we get that $xf(x+y) <f(xf(x))$ meaning that for large enough $x$, $f(xf(x))>k$, and thus resulting in $xf(x)<1$, which contradicts.
Now I will show that $xf(x) \le k$, this is true as $xf(x+y) <f(xf(x)) \le k$, and fixing $x+y$ and varying $y$ to be very small we conclude.
Let's check that $f$ is injective. Assume otherwise and let $f(y_1)=f(y_2)$ and $y_1-y_2=c; c>0$. $P(x;y_1)-P(x;y_2) \Rightarrow f(x)=f(x+c) \forall x \ge y_1$. However this means that $f(x)=f(x+nc) \le \frac{k}{x+nc}$ for natural $n$, which again can't be.
$P(1;y)-P(1;f(y)) \Rightarrow f(f(y))=y, \; \forall y$, because of injectivity.
Now assume that $f(xf(x))=k-a$, $a>0$ for some $x$ and $a$. Than $f(xf(y)+1) <k-a$, meaning $f(f(xf(y))+1)>a$, however as $f$ is surjective, and $f(x)\le \frac{k}{x}$, we conclude that it is impossible for $a$ to be positive. Now we know $f(xf(x))=k=f(f(1))$ thus by injectivity that $f(x)=\frac{f(1)}{x}$, and by substituting back we get $f(x)=\frac{1}{x}$, for any positive real $x$ which indeed is a solution.
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megarnie
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The only solution is $\boxed{f(x) = \frac 1x }$, which works. Now we show it's the only solution. Let $P(x,y)$ be the given assertion.

Claim 1: For $x \ge 1$, we have $f(x) \le 1$.
Proof: Suppose for some $x \ge 1$ that $f(x) > 1$.

$P(x, xf(x) - x)$ gives that $xf(xf(x)) <f(xf(x)) \implies x < 1$, absurd. $\square$

Now let $(1)$ denote the resulting inequality\[ f(xf(x)) \le xf(x+y) + 1\]Claim 2: $f$ is injective
Proof: Suppose we had $f(a) \ne f(b)$ for some $a < b$ with $b - a = d$.

$P(x,a)$ compared with $P(x,b)$ gives that $f(x+a) = f(x+b)$ for all $x > 0$, so $f(x) = f(x+d) \forall x  \ge a$.

Setting $y  = d$ for $x  \ge a$ gives that $f(xf(x)) = xf(x) + f(xf(d) + 1) > xf(x)$. This implies that $xf(x) < 1$, so $f(x) < \frac 1x \forall x \ge a$. Thus, $f(x) \to 0$ as $x \to \infty$.

Now, for any fixed $x>0$, taking $y$ sufficiently large in $(1)$ (so that $f(x+y)$ is sufficiently small) gives that $f(xf(x)) \le 1$ for any positive real $x$. However, we also have $f(xf(x)) > xf(x+y)$, so $xf(x+y) < 1$. Since $f(x) = f(x+nd) \forall n \in \mathbb N$, replacing $x$ with $f(x + nd)$ gives that $(x+nd) f(x+y) = 1$. Taking $n$ arbitrarily large gives a contradiction. $\square$

Claim 3: $f$ is an involution
Proof: $P(1,y)$ compared with $P(1, f(y))$ gives\[f(y+1) = f(f(f(y)) + 1)\implies y + 1 = f(f(y)) + 1 \implies f(f(y)) = y. \ \ \ \square\]Claim 4: As $x \to \infty$, $f(x) \to 0$.
Proof: Suppose not and we had $f(x) > c $ for arbitrarily large $x$ (where $c>0$ is some constant).

Now for any fixed $x$ so that $f(x) > c$, choose $y$ so that $f(x+y) > c$. We have $f(xf(x)) > cx$. Now choose $x$ large enough so that $cx > 1$. We get that $f(xf(x)) > 1 \implies xf(x) < 1$, so $c < f(x) < \frac 1x$, which means $cx < 1$, absurd. $\square$

Claim 5: For every constant $c < 1$, there exists $x$ with $x > 1$ and $f(x) > c$.
Proof: $P(1,y): f(y+1) + f(f(y) + 1) = 1$. Choosing $y$ large enough so that $f(y+1) < 1 - c$ gives the desired result. $\square$

Claim 6: $xf(x) \le 1$ and $f(xf(x)) \ge 1$ for all positive reals $x$.
Proof: $P(x, f(y)): xf(x + f(y)) + f(xy + 1) = f(xf(x))$, so $f(xf(x)) > f(xy + 1)$. Setting $y \to \frac yx$ gives\[ f(xf(x)) > f(y+1)\]If $f(xf(x)) < 1$, then choosing $y$ so that $f(y+1) > f(xf(x))$ (from claim 5) gives a contradiction. Therefore, $f(xf(x)) \ge 1$ for all positive reals $x$. It remains to show that $xf(x) \le 1$ for all $x > 0$. Suppose for some $x$ that $xf(x) > 1$. Since $f(xf(x)) \le 1$, we have $f(xf(x)) = 1$, so $xf(x) = f(1)$, absurd since $f(1) \le 1$. $\square$

We now have\[ f(xf(x)) \le \frac{x}{x+y} \cdot (x+y) f(x+y) + 1 \le 1 + \frac{x}{x + y}\]Setting $y$ arbitrarily large gives a contradiction if $f(xf(x)) > 1$, so $f(xf(x)) = 1$ for all positive reals $x$. Since $f$ is injective, we have $xf(x) = f(1)$ for all $x$, so $f(x) = \frac{f(1)}{x}$.

Claim 7: $f(1) = 1$.
Proof: Note that if not, then $f(1) < 1$. We then have for all $x > 1$ that $f(x) = \frac{f(1)}{x} < f(1)$. However, by claim 5 we can choose $x$ with $x > 1$ and $f(x) > f(1)$, a contradiction. $\square$

Thus $f(x) = \frac 1x$ must hold.
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