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

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0 replies
jlacosta
Mar 2, 2025
0 replies
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
Wait wasn't it the reciprocal in the paper?
Supercali   6
N 27 minutes ago by kes0716
Source: India TST 2023 Day 2 P1
Let $\mathbb{Z}_{\ge 0}$ be the set of non-negative integers and $\mathbb{R}^+$ be the set of positive real numbers. Let $f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \ge 0$, and $$f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1)}{f(m-1, n)+f(m, n-1)}$$for all integers $m, n \ge 1$. Prove that $f(99, 99)<1.99$.

Proposed by Navilarekallu Tejaswi
6 replies
+1 w
Supercali
Jul 9, 2023
kes0716
27 minutes ago
Inspired by Kazakhstan 2017
sqing   0
38 minutes ago
Source: Own
Let $a,b,c\ge \frac{1}{2}$ and $a+b+c=2. $ Prove that
$$\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge 1$$Let $a,b,c\ge \frac{1}{3}$ and $a+b+c=1. $ Prove that
$$\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge 9$$
0 replies
1 viewing
sqing
38 minutes ago
0 replies
About old Inequality
perfect_square   0
44 minutes ago
Source: Arqady
This is: $a,b,c>0$ which satisfy $abc=1$
Prove that: $ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$
By $  uvw $ method, I can assum $b=c=x,a=\frac{1}{x^2}$
But I can't prove:
$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $
Is there an another way?
0 replies
perfect_square
44 minutes ago
0 replies
inquality
karasuno   1
N an hour ago by sqing
The real numbers $x,y,z \ge \frac{1}{2}$ are given such that $x^{2}+y^{2}+z^{2}=1$. Prove the inequality $$(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2 .$$
1 reply
karasuno
2 hours ago
sqing
an hour ago
Number Theory
karasuno   0
2 hours ago
Solve the equation $$n!+10^{2014}=m^{4}$$in natural numbers m and n.
0 replies
karasuno
2 hours ago
0 replies
Minimum number of values in the union of sets
bnumbertheory   5
N 2 hours ago by Rohit-2006
Source: Simon Marais Mathematics Competition 2023 Paper A Problem 3
For each positive integer $n$, let $f(n)$ denote the smallest possible value of $$|A_1 \cup A_2 \cup \dots \cup A_n|$$where $A_1, A_2, A_3 \dots A_n$ are sets such that $A_i \not\subseteq A_j$ and $|A_i| \neq |A_j|$ whenever $i \neq j$. Determine $f(n)$ for each positive integer $n$.
5 replies
bnumbertheory
Oct 14, 2023
Rohit-2006
2 hours ago
two triangles have equal circumradii
littletush   5
N 3 hours ago by Taha.kh
Source: Italy TST 2009 p5
Two circles $O_1$ and $O_2$ intersect at $M,N$. The common tangent line nearer to $M$ of the two circles touches $O_1,O_2$ at $A,B$ respectively. Let $C,D$ be the symmetric points of $A,B$ with respect to $M$ respectively. The circumcircle of triangle $DCM$ intersects circles $O_1$ and $O_2$ at points $E,F$ respectively which are distinct from $M$. Prove that the circumradii of the triangles $MEF$ and $NEF$ are equal.
5 replies
littletush
Mar 10, 2012
Taha.kh
3 hours ago
Equal lengths in cyclic quadrilateral
LoloChen   4
N 3 hours ago by Nari_Tom
Source: All-Russian MO 2024 9.4
In cyclic quadrilateral $ABCD$, $\angle A+ \angle D=\frac{\pi}{2}$. $AC$ intersects $BD$ at ${E}$. A line ${l}$ cuts segment $AB, CD, AE, DE$ at $X, Y, Z, T$ respectively. If $AZ=CE$ and $BE=DT$, prove that the diameter of the circumcircle of $\triangle EZT$ equals $XY$.
4 replies
LoloChen
Apr 22, 2024
Nari_Tom
3 hours ago
Two circles and many points
CHN_Lucas   5
N 4 hours ago by Captainscrubz
Source: 2022 China Second Round A2
$A,B,C,D,E$ are points on a circle $\omega$, satisfying $AB=BD$, $BC=CE$. $AC$ meets $BE$ at $P$. $Q$ is on $DE$ such that $BE//AQ$. Suppose $\odot(APQ)$ intersects $\omega$ again at $T$. $A'$ is the reflection of $A$ wrt $BC$. Prove that $A'BPT$ lies on the same circle.
5 replies
CHN_Lucas
Dec 22, 2022
Captainscrubz
4 hours ago
Angle QRP = 90°
orl   12
N 4 hours ago by YaoAOPS
Source: IMO ShortList, Netherlands 1, IMO 1975, Day 1, Problem 3
In the plane of a triangle $ABC,$ in its exterior$,$ we draw the triangles $ABR, BCP, CAQ$ so that $\angle PBC = \angle CAQ = 45^{\circ}$, $\angle BCP = \angle QCA = 30^{\circ}$, $\angle ABR = \angle RAB = 15^{\circ}$.

Prove that

a.) $\angle QRP = 90\,^{\circ},$ and

b.) $QR = RP.$
12 replies
orl
Nov 12, 2005
YaoAOPS
4 hours ago
IMO 2014 Problem 4
ipaper   166
N 4 hours ago by hgomamogh
Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$. Let $M$ and $N$ be the points on $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$. Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$.

Proposed by Giorgi Arabidze, Georgia.
166 replies
ipaper
Jul 9, 2014
hgomamogh
4 hours ago
IMO 2016 Problem 1
quangminhltv99   146
N 4 hours ago by Ilikeminecraft
Source: IMO 2016
Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.
146 replies
quangminhltv99
Jul 11, 2016
Ilikeminecraft
4 hours ago
A function equation
YaWNeeT   8
N 5 hours ago by HamstPan38825
Source: 2017 Taiwan TST 2nd round day 2 P4
Find all integer $c\in\{0,1,...,2016\}$ such that the number of $f:\mathbb{Z}\rightarrow\{0,1,...,2016\}$ which satisfy the following condition is minimal:
(1) $f$ has periodic $2017$
(2) $f(f(x)+f(y)+1)-f(f(x)+f(y))\equiv c\pmod{2017}$

Proposed by William Chao
8 replies
YaWNeeT
Apr 15, 2017
HamstPan38825
5 hours ago
Circumcenter lies on altitude
ABCDE   58
N 5 hours ago by cj13609517288
Source: 2016 ELMO Problem 2
Oscar is drawing diagrams with trash can lids and sticks. He draws a triangle $ABC$ and a point $D$ such that $DB$ and $DC$ are tangent to the circumcircle of $ABC$. Let $B'$ be the reflection of $B$ over $AC$ and $C'$ be the reflection of $C$ over $AB$. If $O$ is the circumcenter of $DB'C'$, help Oscar prove that $AO$ is perpendicular to $BC$.

James Lin
58 replies
ABCDE
Jun 24, 2016
cj13609517288
5 hours 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
1448 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
1448 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
1482 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
4999 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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