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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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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
gcd nt from switzerland
AshAuktober   0
6 minutes ago
Source: Swiss 2025 Second Round
Let $a, b$ be positive integers. Prove that the expression
\[\frac{\gcd(a+b,ab)}{\gcd(a,b)}\]is always a positive integer, and determine all possible values it can take.
0 replies
AshAuktober
6 minutes ago
0 replies
set construction nt
top1vien   2
N 10 minutes ago by top1vien
Is there a set of 2025 positive integers $S$ that satisfies: for all different $a,b,c,d\in S$, we have $\gcd(ab+1000,cd+1000)=1$?
2 replies
1 viewing
top1vien
Yesterday at 10:04 AM
top1vien
10 minutes ago
strange geometry problem
Zavyk09   0
13 minutes ago
Source: own
Let $ABC$ be a triangle with circumcenter $O$ and internal bisector $AD$. Let $AD$ cuts $(O)$ again at $M$ and $MO$ cuts $(O)$ again at $N$. Point $L$ lie on $AD$ such that $(AD, LM) = -1$. The line pass through $L$ and perpendicular to $AD$ intersects $NC, NB$ at $P, Q$ respectively. Let circumcircle of $\triangle NPQ$ cuts $(O)$ at $G \ne N$. Prove that $\angle AGD = 90^{\circ}$.
0 replies
Zavyk09
13 minutes ago
0 replies
A sharp one with 3 var (3)
mihaig   3
N 16 minutes ago by JARP091
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a^2+b^2+c^2+5abc\geq8.$$
3 replies
mihaig
Yesterday at 5:17 PM
JARP091
16 minutes ago
Dophantine equation
MENELAUSS   2
N 17 minutes ago by Assassino9931
Solve for $x;y \in \mathbb{Z}$ the following equation :
$$3^x-8^y =2xy+1 $$
2 replies
MENELAUSS
Yesterday at 11:35 PM
Assassino9931
17 minutes ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   11
N 20 minutes ago by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
11 replies
OgnjenTesic
May 22, 2025
JARP091
20 minutes ago
Shortest number theory you might've seen in your life
AlperenINAN   9
N 20 minutes ago by Assassino9931
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
9 replies
AlperenINAN
May 11, 2025
Assassino9931
20 minutes ago
Inequality about number of spanning trees of graph
CBMaster   0
39 minutes ago
Let \( k(G) \) be the number of spanning trees in a graph \( G \), where \( G \) may have multiple edges and loops.

For two edges \( e \) and \( f \) of \( G \), let \( G/e \), \( G/f \), and \( G/\{e,f\} \) denote the graphs obtained by contracting the edges \( e \), \( f \), and both \( e \) and \( f \) in $G$, respectively.

Find a combinatorial proof of the following inequality:
\[
k(G/\{e,f\}) \cdot k(G) \leq k(G/e) \cdot k(G/f)
\]
0 replies
CBMaster
39 minutes ago
0 replies
1,2,...,2011 around circle such that 8 of 25 successive multiples of 5 and/or 7
parmenides51   1
N an hour ago by ririgggg
Source: 2011 Belarus TST 2.1
Is it possible to arrange the numbers $1,2,...,2011$ over the circle in some order so that among any $25$ successive numbers at least $8$ numbers are multiplies of $5$ or $7$ (or both $5$ and $7$) ?

I. Gorodnin
1 reply
parmenides51
Nov 8, 2020
ririgggg
an hour ago
Sipnayan JHS 2021 F-9
PikaVee   1
N an hour ago by PikaVee
Matt and Sai are playing a game of darts together. Matt has a slightly more accurate aim than Sai. In
fact, Matt can hit the bullseye 80% of the time while Sai can only hit it 60% of the time. They take turns
in playing and the first player is determined by a flip of a fair coin. If the probability that Sai scores the
first bullseye is given by $ \frac {a}{b} $ where a and b are relatively prime integers, what is b − a?
1 reply
PikaVee
an hour ago
PikaVee
an hour ago
Standart looking FE
Kimchiks926   13
N an hour ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 5
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(0)+1=f(1)$ and for any real numbers $x$ and $y$,
$$ f(xy-x)+f(x+f(y))=yf(x)+3 $$
13 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
an hour ago
A sharp one with 3 var (2)
mihaig   4
N an hour ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a+b+c+\sqrt{abc}\geq4.$$
4 replies
mihaig
May 26, 2025
mihaig
an hour ago
3 var inequality
SunnyEvan   11
N an hour ago by mihaig
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
11 replies
SunnyEvan
May 17, 2025
mihaig
an hour ago
trigonometric inequality
MATH1945   12
N an hour ago by mihaig
Source: ?
In triangle $ABC$, prove that $$sin^2(A)+sin^2(B)+sin^2(C) \leq \frac{9}{4}$$
12 replies
MATH1945
May 26, 2016
mihaig
an hour ago
another functional inequality?
Scilyse   32
N May 5, 2025 by ihategeo_1969
Source: 2023 ISL A4
Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\]for every $x, y \in \mathbb R_{>0}$.
32 replies
Scilyse
Jul 17, 2024
ihategeo_1969
May 5, 2025
another functional inequality?
G H J
Source: 2023 ISL A4
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Scilyse
387 posts
#1 • 5 Y
Y by GrantStar, ehuseyinyigit, OronSH, ohiorizzler1434, NCbutAN
Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\]for every $x, y \in \mathbb R_{>0}$.
This post has been edited 4 times. Last edited by Scilyse, Feb 11, 2025, 9:51 AM
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Math-48
44 posts
#2 • 4 Y
Y by OronSH, ZVFrozel, BorivojeGuzic123, Muaaz.SY
Let $P(x,y)$ denote the assertion:
$$x(f(x)+f(y))\geq (f(f(x))+y)f(y)$$$$P(f(x),x)+P(f(f(x)),f(x)): 2f(f(x))\geq f(f(f(f(x))))+x$$Easy induction gives$f(f(x))> \frac{n}{n+1} x$

Now take $n\to +\infty ~$ to get $~f(f(x))\geq x$

Using this inequality in $P(x,y)$ gives us:

$$xf(x)\geq yf(y)\implies xf(x)=c\implies f(x)=\frac{c}{x}~ \forall x\in \mathbb{R_+}$$Which is indeed a solution for any $c>0$.$~\blacksquare$
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OronSH
1748 posts
#3 • 4 Y
Y by peace09, megarnie, scannose, MS_asdfgzxcvb
The answer is $f(x)=\frac cx$ for constant $c,$ which clearly works.

First taking $x=y$ gives $f(f(x))\le x.$

Now rearrange the inequality to $xf(x)\ge f(y)(f(f(x))+y-x).$ This becomes \[\frac{xf(x)}{yf(y)}\ge1+\frac{f(f(x))-x}y.\]Now the RHS gets arbitrarily close to $1$ from below (or equals $1$) as $y$ gets large, for fixed $x.$ Additionally this gives us that as $y$ gets large, $f(y)$ gets arbitrarily close to $0.$

Then setting $x=f(y)$ in the original inequality gives $y-f(f(y))\le f(y)-f(f(f(y)))<f(y),$ so $y-f(f(y))$ gets arbitrarily close to $0$ for large $y.$

Now if we swap variable names, we get \[\frac{yf(y)}{xf(x)}\ge 1+\frac{f(f(y))-y}x.\]If we fix $x$ and make $y$ arbitrarily large, the RHS gets arbitrarily close to $1$ from below.

Now we have arbitrarily tight bounds on $\frac{xf(x)}{yf(y)}$ in both directions. Thus fixing some positive reals $x,z$ gives us $\frac{xf(x)}{yf(y)}$ and $\frac{yf(y)}{zf(z)}$ are arbitrarily close to $1,$ so their product, which is constant, must be $1.$ Thus $xf(x)$ is some constant $c$ for all $x,$ and thus $f(x)=\frac cx.$
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Aiden-1089
300 posts
#4 • 2 Y
Y by OronSH, Sourena-Majedi
Solving this during TST was an extremely proud moment for me.

Let $P(x,y)$ denote the assertion $x \cdot (f(x)+f(y)) \geq (f(f(x))+y) \cdot f(y)$.
$P(x,x) \implies 2xf(x) \geq f(x)f(f(x))+xf(x) \implies x \geq f(f(x))$.
$P(f(x),x) \implies f(f(x))+f(x) \geq f^3(x)+x \implies f(x)-f^3(x) \geq x-f(f(x))$.

Now assume $f(f(a))>a$ for some $a \in \mathbb{R}^+$.
Then put $a-f(f(a))=c$, note that for all positive integers $n$, we may inductively show that $f^{2n}(a)-f^{2n+2}(a) \geq c$.
But this would imply that there exists some $n$ such that $f^n(a)<0$, contradiction.
Hence $f(f(x))=x$ for all $x$.

Putting this back into the original equation, we get $x \cdot (f(x)+f(y)) \geq (x+y) \cdot f(y) \implies xf(x) \geq yf(y)$ for all $x,y$.
So $xf(x)=yf(y)$ for all $x,y$. Put $xf(x)=k$ for some constant $k$, we see that $f(x)=\frac{k}{x}$. It is easy to check that this is a solution.

Hence the solutions are $f(x)=\frac{k}{x}$ for some constant $k \in \mathbb{R}^+$.
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Sunshine132
136 posts
#5 • 1 Y
Y by OronSH
We take $P(f(x), x)$ to get $f(x)(f(f(x))+f(x)) \geq f(x)(f(f(f(x))) + x) \iff f(f(x)) + f(x) \geq f(f(f(x))) + x$
Thus $f(x) + f^2(x) \geq f^3(x) + x$. Take $x$ to $f(x) \implies f^2(x) + f^3(x) \geq f^4(x) + f(x)$. Adding them, we get $2 f^2(x) \geq f^4(x) + x$.
Easy induction to get: $f^{2n}(x) \leq x + n(f^2(x) - x)$

If $f^2(x) < x \implies f^{2n}(x) < 0$, for $n$ large enough.
So, $f^2(x) \geq x$.

$P(x, x) \implies 2x f(x) \geq f(x) (x + f^2(x)) \iff 2x \leq x + f^2(x) \implies f(f(x)) = x$

$P(x, y) \implies x f(x) \leq y f(y)$. Swapping $x$ and $y$, we get $f(x) = \frac{c}{x}$, for some constant $c$. This function clearly works.
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megarnie
5611 posts
#6 • 1 Y
Y by OronSH
The only solutions are $f(x) = \frac{c}{x}$ for some positive real constant $c$. These clearly work. Now we prove they are the only solutions.

Let $P(x,y)$ denote the given assertion.

$P(x, x): 2xf(x) \ge (f(f(x)) + x) f(x) \implies f(f(x)) \le x$.

$P(f(x), x): f(x) (f(f(x)) + f(x)) \ge (f^3(x) + x)f(x)$, so $f(f(x)) + f(x) \ge f^3(x) + x$, implying that $f(x) - f^3(x) \ge x - f(f(x))$.

Claim: $f$ is an involution
Proof: Consider some $x$ where $f(f(x)) < x$. Then let $d = x - f(f(x)) > 0$. We have $d \le f(x) - f^3(x) \le f(f(x)) - f^4(x) \ldots, $ so $d \le f^n(x) - f^{n+2}(x)$ for any nonnegative integer $n$, meaning that $f^{n+2}(x) \le f^n(x) - d$. Next we induct to show that \[f^{2n}(x) \le x - n\cdot d\]for any positive integer $n$. The base case $n = 1$ holds from $f(f(x)) = x - d$. Suppose it was true for $2k$. Then we have $f^{2k + 2} (x) \le f^{2k}(x) - d \le x - k \cdot d - d = x - (k+1)d$, as desired.

Since $d > 0$, choose $n$ sufficiently large so that $x - n \cdot d < 0$. Then $f$ becomes negative, absurd. $\square$

Now the equation becomes $x(f(x) + f(y)) \ge (x + y) f(y)$, so $xf(x) \ge y f(y)$ and since $y f(y) \ge xf(x)$ also by swapping $x,y$, we have that $xf(x) = yf(y)$ for all reals $x,y$, so $xf(x) = f(1)$, meaning $f(x) = \frac{f(1)}{x}$, as desired.
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Marinchoo
407 posts
#7 • 1 Y
Y by isomoBela
As usual, $P(x,y)$ denotes the assertion of $(x,y)$ into the functional inequality, and $f^k(x)$ denotes an iteration rather than a power. First, $P(x,x)$ yields $x \geq f(f(x))$, and then $P(f(x),x)$ gives
\[f(x)(f^2(x)+f(x)) \geq (f^3(x)+x)f(x)\Longrightarrow f^2(x) + f(x) \geq f^3(x) + x.\]Plugging $f(x)$ in the last inequality and summing:
\begin{align*}
                f^3(x) + f^2(x) &\geq f^4(x) + f(x)\\
                f^2(x) + f(x) &\geq f^3(x) + x\\
\Longrightarrow (f^3(x) + f^2(x)) + (f^2(x) + f(x)) &\geq (f^4(x) + f(x)) + (f^3(x) + x)\\
\Longrightarrow f^2(x) - f^4(x) &\geq x - f^2(x).
\end{align*}Hence, if $a_k(x) = f^{2k}(x) - f^{2k+2}(x)$, we know from before that $a_0(x) \geq 0$, but if $a_0(x)>0$ for some $x$, then from the last inequality, the sequence $a_k(x)$ is monotonically increasing. However, this implies $f^{2k}(x) \leq x - k(x-f^2(x))$, which for large enough $k$ becomes negative, contradiction.

Therefore, $f^2(x) = x$ for all $x>0$, and the original functional inequality becomes simply $xf(x) \geq yf(y)$. This is possible only if $xf(x)$ is constant, i.e. $f(x) = \frac{c}{x}$ for all $x$ and some positive constant $c$. These functions clearly work, so we're done.
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Tqhoud
26 posts
#8
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let

$$P(x,y):x(f(x)+f(y))\ge (ff(x)+y)f(y)$$
$$P(x,x):x\ge ff(x)$$
$$P(f(y),y):ff(y)+f(y)\ge fff(y)+y$$
$$P(ff(y),f(y)):fff(y)+ff(y)\ge ffff(y)+f(y)$$
So



$$2ff(x)\ge y+ffff(y)$$
$$ff(y)-ffff(y)\ge y-ff(y)$$
if exist a number $y$ such that

$$y>ff(y)$$

We can define $ C \colon \mathbb N_{>0} \to \mathbb R_{\ge 0}$ is a function such that

$$f^{2i-2}(y)=f^{2i}(y)+C(1)+C(2)+.....+C(i-1)+C(i)$$
and$C(1)>0$ because $y=ff(y)+C(1)$

it appears that $f^{2i-2}(y)-C(1)\ge f^{2i}(y)$

and because $C(1)$ is a constant there is a positive integer $k$ such that

$$f^{2k}(y)<0$$
wrong result

So $C(1)=0$ and $y=ff(y)$

by editing in the main inequality

$$xf(x)\ge yf(y)$$
and this lead to be $f(x)=\frac{c}{x}$. where $c$ is a positive real number and these functions work
This post has been edited 8 times. Last edited by Tqhoud, Jul 18, 2024, 8:57 AM
Reason: .
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pi271828
3371 posts
#9 • 2 Y
Y by peace09, Funcshun840
The answer is $f(x) \equiv \tfrac{c}{x}$ where $c > 0$. Let $P(x,y)$ denote the given assertion. We have that $P(x,x)$ gives $f(f(x)) \le x $ and $P(f(x), x)$ gives $x - f(f(x)) \le f(x) - f(f(f(x)))$. This implies that \[x - f^2(x) \le f^2(x) - f^4(x) \le f^4(x) - f^6(x) \cdots\]
Claim: $f(f(x)) = x$ for all $x$

Assume for contradiction that there exists an $x$ such that $x > f(f(x))$. Then the sequence $f^{2k}(x)$ will always skip down by at least $x-f(f(x))$ as $k$ increments. Therefore, $f^{2k}(x)$ will be negative for sufficiently large enough $k$, resulting in the desired contradiction.

We can easily plug in the previous claim into the assertion to get $xf(x) \ge yf(y)$ and from swapping we obviously get $yf(y) \ge xf(x)$, implying $xf(x)$ is constant. Therefore we have $f(x) \equiv \tfrac{c}{x}$ which clearly works.
This post has been edited 2 times. Last edited by pi271828, Jul 17, 2024, 6:07 PM
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dkedu
180 posts
#11 • 1 Y
Y by BorivojeGuzic123
We claim the only solution is $f(x) =  \frac cx$ for positive $c$ which is easy to see that these work.

Let $P(x,y)$ denote the assertion. By taking $P(f(y),y)$, we get that
\[y - f(f(y)) \le f(f(f(y))) - f(y)\]But this means
\[y - f(f(y)) \le f^4(y) - f^2(y) \cdots\]So we we have $f(f(y)) = y$, otherwise repeat the above pattern until we get a $f^{2n}(y) < 0$ which is a contradiction. Plug this back in to get
\[x(f(x) + f(y)) \ge (x+y)f(y) \implies xf(x) \ge yf(y) \implies xf(x) = yf(y) \: \forall x,y \in \mathbb{R}_{>0}\]So we recall our original solutions.
This post has been edited 1 time. Last edited by dkedu, Jul 17, 2024, 11:13 PM
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Sammy27
83 posts
#12 • 1 Y
Y by Eka01
Solution
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Yue-Zhang_3906
15 posts
#13
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Easy problem :D
$P(x,x): x \ge f(f(x))$
$P(f(y),y): f(y)-f(f(f(y))) \ge y-f(f(y))$
By repeat iteration, we get sequence ${f^{2k+2}(y)-f^{2k}(y)}$ does not decrease
But when $y-f(f(y))>0$, this will lead to $f^{2N}(y)<0$ with sufficiently large N by summing up the differences
So $y=f(f(y))$ then we can easily know $xf(x)=yf(y)$ for all positive real number $x,y$, and this lead to $f(x)=\frac{c}{x}$
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shanelin-sigma
167 posts
#14
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Am I right?

https://cdn.discordapp.com/attachments/1139915557987160114/1264233135781183510/SPOILER_IMG_2219_2.jpeg?ex=669d205b&is=669bcedb&hm=e73302bc768e30e5e650542c9b438debcd1e8a1a1b6a2ab523240b2f79465c22&

remark
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brainfertilzer
1831 posts
#15
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First, take $y = x$ to get $x\ge f^2(x)$. Now fix a real $a_0 > 0$ and define the sequence $a_n = f^n(a_0)$ for each $n\ge 0$. We claim the following:

Claim 1: $a_{n+1} - a_{n+3}\ge a_{n} - a_{n+2}$.
proof: Take $(x,y) = (a_{n+1}, a_n)$ in the FE to get $a_{n+1}a_{n+2} + a_{n+1}^2\ge a_{n+3}a_{n+1}+ a_na_{n+1}$. Divide by $a_{n+1}$ and rearrange to finish $\square$.

Claim 2: $a_{2n}\le a_0 - (a_0-a_2)n$
proof: Induct on $n$. Base case of $n = 0$ is obvious, now suppose it holds for some $n$. Then
\[ a_{2n+2}\le 2a_{2n} - a_{2n-2}  = a_{2n} -(a_{2n-2} - a_{2n}) \le a_{2n} - (a_0 - a_2)\le a_0 - (a_0-a_2)(n+1),\]done $\square$

A consequence of claim 2 is that $\{a_{2n}\}_{n\ge 0}$ is unbounded from below if $a_0-a_2 > 0$. But that's impossible since $\{a_{2n}\}_{n\ge 0}$ is a sequence of positive reals. Since $a_0 - a_2\ge 0$, we are then forced to have $a_0 - a_2 = 0$. Hence $a_0 = a_2$, meaning $f(f(a_0)) = a_0$ for all $a_0$. The original FE then rewrites as
\[ xf(x) + xf(y)\ge xf(y) + yf(y)\implies xf(x)\ge yf(y)\]for all $x,y > 0$. Swap $x,y$ to get $xf(x) = yf(y)$ for all $x,y$. Take $y = 1$ to finally get $\boxed{f(x) = f(1)/x}$ for all $x$, which clearly works.
This post has been edited 1 time. Last edited by brainfertilzer, Jul 20, 2024, 8:52 PM
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VicKmath7
1391 posts
#16
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Solution
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Pyramix
419 posts
#17
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Given condition is equivalent to:
$P(x,y):\ \left(f\left(f\left(x\right)\right)+y-x\right)f\left(y\right)\le xf\left(x\right)$
$P(x,x):\ f(f(x))\leq x$.

Claim: $f(f(x))=x$ for every $x>0$.
Proof. $P(f(x),x):\ 0\leq x-f(f(x)) \leq f(x) - f(f(f(x)))$
Hence, if $t=x-f(f(x))>0$, then $f^n(x)-f^{n+2}(x)\geq t$ for every $n\geq 0$. So, for any $n$, we have \[f^{2n}(x)\geq t\Longrightarrow f^{2n-2}(x)\geq 2t, f^{2n-4}(t)\geq 3t, \ldots, x\geq nt,\]which is impossible for sufficiently large $n$. This forces $t=0$, and $f(f(x))=x$ for every $x>0$. $\blacksquare$

This means $P(x,y)\Longleftrightarrow xf(x)\geq yf(y)$, while $P(y,x)\Longleftrightarrow yf(y)\geq xf(x)\Longrightarrow xf(x)=yf(y)$. So, $xf(x)=c$ for some constant $c>0$. So, $f(x)=\frac cx$ for all $x$, which indeed satisfies the original condition. $\blacksquare$
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sami1618
916 posts
#18
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The only solution is $f(x)=\frac{c}{x}, c\in \mathbb{R}_{>0}$, which turns the inequality into an equality. Let $P(x,y)$ be the given assertion.

Checking $P(x,f(x))$ gives that $x\geq f(f(x))$. Define $d(x)=x-f(f(x))$. The assertion $P(f(x),x)$ gives that $d(f(x))\geq d(x)$. This implies that $d(f(f(x)))\geq d(x)$. However we have that $$(k+1)d(x)\leq d(x)+d(f^2(x))+\dots+d(f^{2k}(x))=x-f^{2k+2}(x)\leq x$$By taking $k\rightarrow \infty$ we must have that $d(x)=0$, that is $f(f(x))=x$.

This simplifies the assertion to $xf(x)\geq yf(y)$. Then however both sides must be constant, implying the above solution.
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Yue-Zhang_3906
15 posts
#19
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shanelin-sigma wrote:
Am I right?

https://cdn.discordapp.com/attachments/1139915557987160114/1264233135781183510/SPOILER_IMG_2219_2.jpeg?ex=669d205b&is=669bcedb&hm=e73302bc768e30e5e650542c9b438debcd1e8a1a1b6a2ab523240b2f79465c22&

Why does $min(G)$ exist?I think you should use the supremum and infimum principle instead of directly setting the minimum value
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shanelin-sigma
167 posts
#20
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Yue-Zhang_3906 wrote:
Why does $min(G)$ exist?I think you should use the supremum and infimum principle instead of directly setting the minimum value

Oh no, I made an elementary mistake. Sorry :stretcher:
but how could I fix it?
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rrrMath
67 posts
#21 • 1 Y
Y by shanelin-sigma
Based on my solution during my country's TST which it appeared in which also seems to be the first proof of this sort on this thread.

Denote the asserion by $P\left(x,y\right)$.
From $P\left(x,x\right)$ we have $f\left(f\left(x\right)\right)\leq x$.
Now take $x\to0$ in $P\left(x,y\right)$ and denote $c_1=\liminf_{x\to0}{xf\left(x\right)}$:
$$0<f\left(f\left(x\right)\right)\leq x\to0\Rightarrow yf\left(y\right)\leq c_1$$So we have a function that is at most it's liminf meaning $\lim_{x\to0}{xf\left(x\right)}=c_1$.
Now $f\left(y\right)\leq\frac{c_1}{y}$ meaning $\lim_{y\to\infty}{f\left(y\right)}=0$ so taking $y\to\infty$ in $P\left(x,y\right)$ and denoting $c_2=\limsup_{y\to\infty}{yf\left(y\right)}$ gives $xf\left(x\right)\geq c_2$ and similarly to earlier, the function is always at least it's limsup meaning $\lim_{x\to\infty}{xf\left(x\right)}=c_2$.
So for now we have:
$$c_1\geq xf\left(x\right)\geq c_2$$Now notice $c_1\leq c_2$ which follows from taking $x\to\infty$ in $f\left(x\right)f\left(f\left(x\right)\right)\leq xf\left(x\right)$ so we have $xf\left(x\right)$ is constant i.e. $f\left(x\right)=\frac{c}{x},c\in\mathbb{R}_{>0}\thinspace\forall x\in\mathbb{R}_{>0}$ which fits.
This post has been edited 4 times. Last edited by rrrMath, Nov 17, 2024, 8:38 PM
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MR_D33R
15 posts
#22
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Firstly, we rewrite the given inequality as $xf(x)\geqslant f(y)(f(f(x))-x+y)$. Notice that $\lim_{x \to \infty}f(x)=0$. To see this, fix $x$ and take $y$ to infinity. The left hand side of the inequality is constant, but the right hand side is asymptotically $yf(y)$, so $f(y)$ must go to 0. Again, by taking $y$ to infinity we have
\[xf(x)\geqslant \limsup_{y \to \infty}yf(y)\implies M:=\inf \{xf(x) | x \in \mathbb{R}_{>0}\}\geqslant \limsup_{y \to \infty}yf(y).\]Plugging in $x=y$ gives $x\geqslant f(f(x))$ for all $x$. Multiplying this by $f(x)$ gives $xf(x)\geqslant f(x)f(f(x))$ and hence
\[M \geqslant \limsup_{x\to \infty}f(x)f(f(x)).\]Setting $x$ to $f(x)$ in the initial inequality and taking $x$ to infinity gives
\[M \geqslant \limsup_{x\to \infty}f(x)f(f(x)) \geqslant yf(y) -f(y)\limsup_{x\to \infty}(f(x)-f(f(f(x)))),\]but since $\lim_{x \to \infty}f(x)=0$ and $f(x)\geqslant f(f(f(x)))$, we get that the last limsup is equal to 0. Finally we get that
\[M\geqslant yf(y)\]for any $y$. By the definition of $M$, we must have an equality for all $y$, so $f(x)=\frac{M}{x}$ for all $x$. This is in fact a solution for any positive $M$.
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SomeonesPenguin
129 posts
#23
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Straightforward enough.

Solution
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jp62
54 posts
#24 • 1 Y
Y by OronSH
\begin{align*}
s&\geq f(f(s))\tag{$P(s,s)$}\\
&\geq\frac{\alpha f(\alpha)}{f(s)}-f(\alpha)\tag{$P(f(s),\alpha)$ dropping the $f(f(x))$ term}\\
&\geq\frac{\alpha f(\alpha)}{\beta f(\beta)}(s-\beta)-f(\alpha)\tag{$P(\beta,s)$ dropping the $f(f(x))$ term}\\
\end{align*}which fails for $\alpha f(\alpha)>\beta f(\beta)$ and sufficiently large $s>\beta$. We conclude.

Comments / Rant
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InterLoop
279 posts
#25
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based problem
solution
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Cali.Math
128 posts
#26
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We uploaded our solution https://calimath.org/pdf/ISL2023-A4.pdf on youtube https://youtu.be/aDr1Ai0uwgw.
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Abidabi
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#28
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We rewrite as $f(y) (x-f^2(x)) \ge yf(y) - xf(x)$. Putting $(x,x)$ gives $x\ge f^2(x) \Rightarrow f^{n} (x) \ge f^{n+2} (x)$. Notice that $$x\ge f^2(x) \ge f^4(x) \ge \ldots > 0.$$This implies that $\lim\limits_{n\to \infty} f^{2n} (x) - f^{2n + 2}(x) = 0$. Putting $(f^{2n}(x),y)$ gives us
$$f(y)(f^{2n}(x)-f^{2n+2}(x))\ge yf(y) - f^{2n}(x)f^{2n+1}(x) \ge yf(y) - xf(x).$$Fixing $x,y$ and taking $n\to\infty$ implies that $0 \ge yf(y)-xf(x)$, thus $xf(x) \equiv c$ for some $c > 0$.
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N3bula
296 posts
#29 • 1 Y
Y by OronSH
Let $P(x, y)$ be the assertion in the question.
\[P(x, x)\]\[x\leq f(f(x))\]Rearrange the inequality to:
\[\frac{xf(x)}{yf(y)}\geq \frac{f(f(x))-x}{y}+1\]Thus we get that $\frac{xf(x)}{yf(y)}$ gets abitrarily close to $1$ from below as we
make $y$ large. By swapping variables we get that $\frac{yf(y)}{xf(x)}$ gets abitrarily
close to $1$ from below as we make $x$ large. So we get that $xf(x)=c$ for some fixed
$c$. Thus the only solutions are $f(x)=\frac{c}{x}$.
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Gelu
6 posts
#30
Y by
it is very easy such that we just try monotonnia of function F ,and get $f(x)=\frac{c}{x}$
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HamstPan38825
8868 posts
#31
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The answer is $f(x) = \frac cx$ for every positive real $c$, which work.

Setting $y = f(x)$ yields \[x(f(x)+f(f(x)) \geq f(f(x))(f(x)+f(f(x)))\]for every $x$, i.e. $x \geq f(f(x))$ for each $x$. I will show that this must actually be an equality:

Claim: $f(f(x)) = x$ for every real number $x$.

Proof: Assume for the sake of contradiction that there exists an $\varepsilon > 0$ such that $f(f(x_0)) < x_0 - \varepsilon$ for some fixed $x_0$. Setting $f(x_0), x_0$ in the original equation,
\[f(x_0)(f(f(x_0)) + f(x_0)) \geq f(x_0)(f(f(f(x_0))) + x_0)\]so in particular \[f(x_0) - f(f(f(x_0))) \geq f(f(x_0)) - f(x_0) > \varepsilon.\]Repeatedly applying this argument, it follows that \[f^{2n-2}(x_0) - f^{2n}(x_0) < \varepsilon\]for each $\varepsilon$. But picking an $n$ such that $n\varepsilon > x_0$, $f^{2n+2}(x_0) < 0$, which is a contradiction. $\blacksquare$

Thus we have \[x(f(x) + f(y)) \geq (x+y)f(y) \iff xf(x) \geq yf(y)\]for every pair of real numbers $(x, y)$. This implies $xf(x) = yf(y)$ for all $(x, y)$, so $xf(x) = c$, and $f(x) = \frac cx$ for some $c > 0$ all work.
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thaiquan2008
6 posts
#33 • 1 Y
Y by OronSH
Denote $P(x,y)$ the assertion.
It follows from $P(x,x)$ that
$$2xf(x)\ge\big(f\big(f(x)\big)+x\big)f(x)\Longleftrightarrow 2x\ge f\big(f(x)\big)+x\Longleftrightarrow f\big(f(x)\big)\le x$$\begin{align}
f(x)f\big(f(x)\big)\le xf\big(f(x)\big).
\end{align}For any fixed $x>0$ and $y>2x-2f\big(f(x)\big)\ge0$, we have
\begin{align*}
yf(y)&=\left(1+\frac{x-f\big(f(x)\big)}{y+f\big(f(x)\big)-x}\right)\left(y+f\big(f(x)\big)-x\right)f(y)\\
&\le\left(1+\frac{x-f\big(f(x)\big)}{y+f\big(f(x)\big)-x}\right)xf(x)\\
&<2xf(x).
\end{align*}So $yf(y)$ remains bounded as $y\to\infty$, this consequently implies $\lim_{y\to\infty}f(y)=0$.
If we have $\lim_{x\to0^+}xf(x)=\infty$, since $f(x)\to0$ as $x\to\infty$ it is true that $f(x)f\big(f(x)\big)\to\infty$ as $x\to\infty$. But $xf(x)$ is bounded from above as $x\to\infty$, and this contradicts with $(1)$ if we let $x$ goes to $\infty$. Hence we may write $L=\liminf_{x\to0^+}xf(x)<\infty$.
An application of $P(x,y)$ shows that
\begin{align*}
yf(y)&=\liminf_{x\to0^+}yf(y)\\
&\le\liminf_{x\to0^+}\big(xf(y)-f\big(f(x)\big)f(y)+xf(x)\big)\\
&\le\liminf_{x\to0^+}\big(xf(y)+xf(x)\big)\\
&=\liminf_{x\to0^+}xf(x)=L.
\end{align*}So $xf(x)\le L$ for any $x>0$. We consequently have $\liminf_{x\to0^+}xf(x)\le\limsup_{x\to0^+}xf(x)\le L$, and so $\liminf_{x\to0^+}xf(x)=\limsup_{x\to0^+}xf(x)=L$ or $\lim_{x\to0^+}xf(x)=L$.
Now we look again at $(1)$. We have $\liminf_{x\to\infty}xf(x)\le\limsup_{x\to\infty}xf(x)\le L$, on the other hand $f(x)\to0$ as $x\to\infty$ so $\lim_{x\to\infty}f(x)f\big(f(x)\big)=L$. We deduce that $L=\liminf_{x\to\infty}f(x)f\big(f(x)\big)\le\liminf_{x\to\infty}xf(x)$, and hence $\lim_{x\to\infty}xf(x)=L$.
We rewrite the given inequality as
$$xf(x)\ge f\big(f(x)\big)f(y)-xf(y)+yf(y).$$Take the limit as $y\to\infty$ yields $xf(x)\ge L$ for $x>0$. We also proved the inequality $xf(x)\le L$ for all $x>0$, so $f(x)\equiv\frac{L}{x}$. Then the inequality turns into an equality, and hence $f(x)\equiv\frac{L}{x}$ satisfies. All solutions of the problem are in the form $f(x)\equiv\frac{c}{x}$, with $c$ being some positive constant.\qed
This post has been edited 1 time. Last edited by thaiquan2008, Feb 21, 2025, 1:13 PM
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Ilikeminecraft
664 posts
#34
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Claim: $f$ is an involution
Proof: Take $(x, x)$ to get $x\geq f^2(x).$ Take $f^2(x) + f(x)\geq x + f^3(x).$ Rearrangement tells us that $f(x) - f^3(x) \geq x - f^2(x)\geq0.$ In particular, $f^k(x) - f^{k + 2}(x)\geq x - f^2(x).$ If we add these up, we get:
\begin{align*}
        y - f^{2k}(x) & = \sum_{n=0}^{k - 1} f^{2n}(y) - f^{2n + 2}(y) \\
        & \geq \sum_{n = 0}^{k - 1}y - f^2(y) \\
        & = k(y - f^2(y)) \geq 0
    \end{align*}However, $y - f^{2k}(x) < y,$ and so $y - f^2(y) \leq \frac yk,$ and by sending $k$ to infinity, we force $f^2(y) = y.$

To conclude, let $f(1) = c.$ Then, take $x = 1, y = 1$ to get $f\equiv \frac cx.$
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Andyexists
8 posts
#35
Y by
Look mom it's pink

Taking $x = y$ in the statement, we get $x(f(x) + f(x)) \geq (f(f(x)) + x)f(x)$, from which $2xf(x) \geq xf(x) + f(x)f(f(x))$, so $xf(x) \geq f(f(x))f(x)$. We can divide by $f(x)$ since $f(x)>0$, and we find $x \geq f(f(x)), \forall x \in \mathbb{R}$. ($\bigstar$)

Next, by taking $x \rightarrow f(y)$ in the statement, we get $f(y)(f(f(y)) + f(y)) \geq (f(f(f(y))) + y)f(y)$, and dividing by $f(y)$ once more, we get $f(f(y)) + f(y) \geq y + f(f(f(y)))$, from where $f(f(y)) - y \geq f(f(f(y))) - f(y), \forall y \in \mathbb{R}$. Fix $y$, and let $a_n = f^{n}(y)$ (that is $f$ applied $n$ times to $y$).

So we have $a_{n+2} - a_n \geq a_{n+3} - a_{n+1}$. Increasing $n$ by $1$, we also get $a_{n+3} - a_{n+1} \geq a_{n+4} - a_{n+2}$. Joining the two inequalities, we have $a_{n+2} - a_n \geq a_{n+4} - a_{n+2}$.

From $\bigstar$, we know that the difference $a_{n+2} - a_n$ is zero or negative. Assume it's negative, and let $a_{2} = a_0 + c$, with $c < 0$. Then $a_4 - a_2 \leq c$, so $a_4 \leq a_2 + c$, so $a_4 \leq a_0 + 2c$. Inductively, we have $a_{2n} \leq a_0 + n \times c$, and since $a_{2n+2} - a_{2n} \leq c$, we have $a_{2n+2} \leq a_0 + (n+1) \times c$.

We know that $f(x) > 0$ always, so if $c<0$, then $a_0 + nc$ is unbounded negatively as $n$ increases, meaning from a certain $N$ higher, $a_0 + nc < 0$, which implies $a_{2n} < 0$, false. So $c=0$, so $f(f(x)) = x$.

Returning to the statement, we find $xf(x) \geq yf(y)$. Swapping $x$ and $y$, we get $yf(y) \geq xf(x)$, and the two inequalities imply $xf(x) = yf(y) = k$, from where $f(x) = k / x$, which verifies for any $k>0$.
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ihategeo_1969
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Let $P(x,y)$ denote the assertion.

$P(x,x)$ gives us that $x \ge f(f(x))$.

Claim: For all $n \in \mathbb N$, we have $f(f(x)) \ge \frac{nx}{n+1}$ and in particular we have $f(f(x)) \ge x$.
Proof: By summing up $P(f(x),x)$ and $P(f(f(x)),f(x))$, we have \[2f^2(x) \ge x+f^4(x)>x \implies f^2(x) > \frac x2\]So base case is complete. Now assume it i true for $n=N$. Then see that \[2f^2(x) \ge x+f^4(x) \ge x+\frac{N}{N+1} f^2(x) \implies f^2(x) \ge \frac{(N+1)x}{N+2}\]And we are done. $\square$

This implies $f(f(x))=x$ and subbing it back in parent equation we have $xf(x) \ge yf(y)$ so $xf(x)$ is some constant (all such functions can be checked to work).

Hence only solutions are \[\boxed{f(x) \equiv \frac cx \text{ } \forall \text{ } x \in \mathbb R_{>0}} \text{ where } c \in \mathbb R_{>0}\]
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