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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.

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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
Max and Min
Butterfly   0
5 minutes ago

Let $a_1,a_2,\cdots,a_n$ be an arrangement of $\{1,2,3,\cdots,n\}$. Find the maximum and minimum values of $$\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}.$$
0 replies
Butterfly
5 minutes ago
0 replies
Cup of Combinatorics
M11100111001Y1R   4
N 10 minutes ago by sami1618
Source: Iran TST 2025 Test 4 Problem 2
There are \( n \) cups labeled \( 1, 2, \dots, n \), where the \( i \)-th cup has capacity \( i \) liters. In total, there are \( n \) liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.

$a)$ Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most \( \frac{4n}{3} \) steps.

$b)$ Prove that in at most \( \frac{5n}{3} \) steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
4 replies
M11100111001Y1R
May 27, 2025
sami1618
10 minutes ago
(c^n+1)/(2^na+b) is an integer for all n
parmenides51   2
N 12 minutes ago by Assassino9931
Source: Ukraine TST 2010 p6
Find all pairs of odd integers $a$ and $b$ for which there exists a natural number$ c$ such that the number $\frac{c^n+1}{2^na+b}$ is integer for all natural $n$.
2 replies
parmenides51
May 4, 2020
Assassino9931
12 minutes ago
Nine point circle + Perpendicularities
YaoAOPS   18
N 36 minutes ago by AndreiVila
Source: 2025 CTST P2
Suppose $\triangle ABC$ has $D$ as the midpoint of $BC$ and orthocenter $H$. Let $P$ be an arbitrary point on the nine point circle of $ABC$. The line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$. The line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$. If $M$ is the midpoint of $AQ$, show that $HX \perp DM$.
18 replies
YaoAOPS
Mar 5, 2025
AndreiVila
36 minutes ago
Inequality conjecture
RainbowNeos   0
an hour ago
Show (or deny) that there exists an absolute constant $C>0$ that, for all $n$ and $n$ positive real numbers $x_i ,1\leq i \leq n$, there is
\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]
0 replies
RainbowNeos
an hour ago
0 replies
inequality 2905
pennypc123456789   0
an hour ago
Consider positive real numbers \( x, y, z \) that satisfy the condition
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3.
\]Find the maximum value of the expression
\[
P = \dfrac{yz}{\sqrt[3]{3y^2z^2+ 3x^2y^2z^2+ x^2z^2 + x^2y^2}}
+ \frac{xz}{\sqrt[3]{3x^2z^2 + 3x^2y^2z^2 + x^2y^2 + y^2z^2}}
+ \frac{xy}{\sqrt[3]{3x^2y^2 + 3x^2y^2z^2 +y^2z^2 + x^2z^2}}.
\]
0 replies
1 viewing
pennypc123456789
an hour ago
0 replies
Inspired by m4thbl3nd3r
sqing   3
N an hour ago by sqing
Source: Own
Let $  a, b,c>0,b+c>a$. Prove that$$\sqrt{\frac{a}{b+c-a}}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)}\geq 1$$$$\frac{a}{b+c-a}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)} \geq  \frac{4\sqrt 2}{3}-1$$
3 replies
1 viewing
sqing
Today at 3:43 AM
sqing
an hour ago
Inspired by qrxz17
sqing   7
N an hour ago by sqing
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
7 replies
1 viewing
sqing
5 hours ago
sqing
an hour ago
Geometry problem
Whatisthepurposeoflife   2
N an hour ago by Whatisthepurposeoflife
Source: Derived from MEMO 2024 I3
Triangle ∆ABC is scalene the circle w that goes through the points A and B intersects AC at E BC at D let the Lines BE and AD intersect at point F. And let the tangents A and B of circle w Intersect at point G.
Prove that C F and G are collinear
2 replies
Whatisthepurposeoflife
Yesterday at 1:45 PM
Whatisthepurposeoflife
an hour ago
A Sequence of +1's and -1's
ike.chen   36
N an hour ago by maromex
Source: ISL 2022/C1
A $\pm 1$-sequence is a sequence of $2022$ numbers $a_1, \ldots, a_{2022},$ each equal to either $+1$ or $-1$. Determine the largest $C$ so that, for any $\pm 1$-sequence, there exists an integer $k$ and indices $1 \le t_1 < \ldots < t_k \le 2022$ so that $t_{i+1} - t_i \le 2$ for all $i$, and $$\left| \sum_{i = 1}^{k} a_{t_i} \right| \ge C.$$
36 replies
ike.chen
Jul 9, 2023
maromex
an hour ago
Basic ideas in junior diophantine equations
Maths_VC   1
N 2 hours ago by grupyorum
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
1 reply
Maths_VC
Tuesday at 7:54 PM
grupyorum
2 hours ago
Inspired by qrxz17
sqing   3
N 2 hours ago by sqing
Source: Own
Let $ a,b,c $ be reals such that $ (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 = 28 $ and $  (a^2+b^2+c^2)^2 =16. $ Find the value of $ a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1).$
3 replies
sqing
5 hours ago
sqing
2 hours ago
Hardest in ARO 2008
discredit   30
N 2 hours ago by Phat_23000245
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
30 replies
discredit
Jun 11, 2008
Phat_23000245
2 hours ago
Find the value
sqing   12
N 2 hours ago by Phat_23000245
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
12 replies
sqing
Jun 22, 2024
Phat_23000245
2 hours ago
R+ FE with arbitrary constant
CyclicISLscelesTrapezoid   25
N Apr 22, 2025 by DeathIsAwe
Source: APMO 2023/4
Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f \colon \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that \[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\]
25 replies
CyclicISLscelesTrapezoid
Jul 5, 2023
DeathIsAwe
Apr 22, 2025
R+ FE with arbitrary constant
G H J
G H BBookmark kLocked kLocked NReply
Source: APMO 2023/4
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CyclicISLscelesTrapezoid
372 posts
#1 • 5 Y
Y by Johnson100, Rounak_iitr, Maths-I25, GeoKing, Funcshun840
Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f \colon \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that \[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\]
This post has been edited 2 times. Last edited by CyclicISLscelesTrapezoid, Jul 5, 2023, 7:24 PM
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IAmTheHazard
5005 posts
#2 • 4 Y
Y by centslordm, Maths-I25, sttsmet, MS_asdfgzxcvb
The answer is $f(x)=2x$ only, which clearly works. Now we prove that this is the only solution.

Let $P(x,y)$ denote the assertion. If $f(y)<2y$ for any $y$, then from $P(x,\tfrac{2y-f(y)}{c})$ we find that $2cx=0$: contradiction. Therefore we may let $f(x)=g(x)+2x$, where $g: \mathbb{R}^+ \to \mathbb{R}_{\geq 0}$, and the functional equation becomes
$$g((c+1)x+g(y)+2y)+2g(y)=g(x+2y).$$Let $k=2+f(1)$. Picking an arbitrary $t>0$, by $P(t,k)$, we find that $g(t+k)\geq g((c+1)t+k+f(k))$, which really just means that we can always find some $t'\geq(c+1)t$ such that $g(t'+k)\leq g(t+k)$.

By repeatedly applying this, we find that there exists some $t''\geq 2t+k$ such that $g(t''+k)\leq g(t+k)$. But then from $P(t''-2t-k,t+k)$, we find that $g((c+1)(t''-2t-k)+g(t+k)+2(t+k))\leq 0$, with equality holding only if $g(t''+k)=g(t+k)=0$. Therefore $g(y)=0$ when $y>k$. But then from $P(k,x)$ where $x>0$ is arbitrary, we find that $g(x)=0$ for all $x$, hence $g \equiv 0$ and we extract the desired solution. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Jul 5, 2023, 9:03 PM
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kred9
1022 posts
#3 • 2 Y
Y by StarLex1, solasky
The answer is $f(x) = 2x$ only. This works.

Notice that equation $(c+1)x+f(y) = x+2y$ can never have solutions in the positive reals, since otherwise $x$ or $c$ would have to be $0$. This implies that $f(y) \ge 2y \phantom{jj} (*)$ for all $y > 0$. We have two cases.

Case 1. $f(t) = 2t$ for some $t > 0$.
By $*$, we have $$f(x+2y) = f((c+1)x + f(y)) - 2cx \ge 2(c+1)x + 2f(y) -2cx\ge 2x + 4y.$$Pick some $(x,y)$ with $x+2y = t$. Then the above inequality is actually an equality, so we have $f(y) = 2y$ and $f((c+1)x + 2y) = 2(c+1)x + 4y$ for all $x+2y = t$. The second equality simplifies to $f(t + cx) = 2t + 2cx$ for all $x < t$. Since $c+1 > 1$, repeating this process implies that $f(x) = 2x$ for all $x \ge t$, since the function $(c+1)^n$ grows without bound.

In particular, $f(2t) = 4t$, so we have $f(y) = 2y$ for all $y < t$ as well, showing that $f(x) = 2x$ for all $x \in \mathbb{R}_{>0}$.

Case 2. $f(t) > 2t$ for all $t>0$. Then let $g(x) = f(x) - 2x$. Clearly the codomain of $g$ is also $\mathbb{R}_{>0}$.

The given equation simplifies to $g(x+2y) = 2g(y) + g((c+1)x+g(y) + 2y)$. This implies $g(x+2y) > 2g(y)$, so for all pairs $(x,y)$ with $x>2y$, we have $g(x) > 2g(y)$.

Let $g(1) = M>0$. Then for all $x > 2$, $g(x) > 2M$. Taking $y = 1$ into the given equation,
\begin{align*}
g(x+2) &= 2M + g(\textcolor{blue}{(c+1)x+M}+2) \\
&= 2M + 2M + g((c+1)(\textcolor{blue}{(c+1)x+M}) +M + 2) \\
&= 2M + 2M + 2M + g((c+1)((c+1)(\textcolor{blue}{(c+1)x+M})+M) +M + 2) \\
&= \cdots,
\end{align*}which is a contradiction as $M > 0$. We are done.
This post has been edited 3 times. Last edited by kred9, Jul 5, 2023, 11:46 PM
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Math-48
44 posts
#4 • 2 Y
Y by Rounak_iitr, jolynefag
Kind of easy P4, anyway, as it's FE here's my solution : $~$ :play_ball:

Let $P(x,y)$ be the assertion :
$$ f((c+1)x+f(y))=f(x+2y)+2cx$$Claim : $f(x)\ge 2x ~\forall x \in \mathbb{R_+}$

Proof : if $\exists w \in \mathbb{R_+}~ ; f(w)<2w :$
$$P(\frac{2w-f(w)}{c},w): f(2w+\frac{2w-f(w)}{c})=f(2w+\frac{2w-f(w)}{c})+2(2w-f(w))$$$\implies f(w)=2w$ contradiction. $\blacksquare$

Define :$$g:\mathbb{R_+} \to \mathbb{R}_{\geq 0} 
 ~,~ g(x)=f(x)-2x\implies f(x)=g(x)+2x$$$\bullet ~P(x,y)$ becomes $:$
$$g((c+1)x+2y+g(y))+2g(y)=g(x+2y)$$$$g((c+1)x+2y+g(y))\ge 0\implies g(x+2y)\ge 2g(y)$$and so $x>2y\implies g(x)\ge 2g(y) $

$\bullet$ Now assume that$: \exists\alpha\in\mathbb{R_+}~ ; g(\alpha)>0$
$$P(x-\frac{2\alpha+g(\alpha)}{c+1},\alpha) ~\text{with}~  
 x>\frac{2\alpha+g(\alpha)}{c+1} : $$$g((c+1)x)+2g(\alpha)=g(x+\frac{2c\alpha-g(\alpha)}{c+1})$
$$a:=c+1>1,r:=\frac{1}{a}<1,t:=2g(\alpha)>0,d:=\frac{2c\alpha-g(\alpha)}{c+1}\in\mathbb{R}$$$$\implies g(ax)+t=g(x+d)\implies g(ax)=g(x+d)-t$$§ Now change $x\to ax:$
$$g(a^2x)=g(ax+d)-t=g(a(x+rd))-t=g(x+rd+d)-2t$$§ again change $x\to ax:$
$$g(a^3x)=g(ax+rd+d)-2t=g(a(x+r^2d+rd))-2t=g(x+r^2d+rd+d)-3t$$§ By induction we get $:$
$$ {g(a^nx)+nt=g(x+d\frac{1-r^n}{1-r}) ~ \forall n \in\mathbb{N}}  $$$\bullet$ Now put $x=x_0$ where $x_0$ is a fixed positive real number satisfying $:$

$x_0+d\frac{1-r^n}{1-r}>0, \forall n\in\mathbb{N}$ which is exist

because $\frac{1-r^n}{1-r}<\frac{1}{1-r}~$ (bounded)

So $x_0+d\frac{1-r^n}{1-r}$ is bounded $\forall n\in\mathbb{N}$
$$\implies \exists l \in\mathbb{R_+}~ ; l>x_0+d\frac{1-r^n}{1-r} ~\forall n\in\mathbb{N}$$$$\implies 2l>2(x_0+d\frac{1-r^n}{1-r})\implies g(2l)\ge 2g(x_0+d\frac{1-r^n}{1-r})~\forall n\in\mathbb{N}$$So $g(x_0+d\frac{1-r^n}{1-r})$ is bounded $\forall n\in\mathbb{N}$

$\bullet$ Now take $n\to +\infty$ in the equation $:$
$$g(a^nx_0)+nt=g(x_0+d\frac{1-r^n}{1-r})$$we see that $RHS$ is bounded, but $LHS$ goes to infinity, so we get a contradiction and so no such $\alpha$ exist.
$$\implies g(x)=0 ~ \forall x\in\mathbb{R_+}\implies \boxed{ f(x)=2x ~ \forall x \in\mathbb{R_+} }  $$remark
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yofro
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#5 • 1 Y
Y by LLL2019
See this thread from 2019. The solution should adapt.
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Orestis_Lignos
558 posts
#6
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The only solution is $f(x)=2x$, which evidently works. From now on, assume that $f$ is a solution which is different from this function. We prove a series of Claims:

Claim 1: $f(x) \geq 2x$ for all $x>0$.
Proof: Indeed, if $f(u)<2u$ for some $u>0$, then we may take $x=\dfrac{2u-f(u)}{c}$ and $y=u$ in the given equation to obtain that $2c \cdot \dfrac{2u-f(u)}{c}=0,$ a contradiction $\blacksquare$

Claim 2: If $x>2y$, then $f(x)-2x \geq 2(f(y)-2y)$.
Proof: Indeed, note that

$f(x+2y)-2(x+2y) =f((c+1)x+f(y))-2cx-2x-4y \geq 2((c+1)x+f(y))-2cx-2x-4y=2(f(y)-2y),$

and so $f(x)-2x \geq 2(f(y)-2y)$ for all $x,y$ such that $x>2y$ $\blacksquare$

To the problem, note that

$(c+1)x+f(y)>f(y)>2y,$

and so

$f((c+1)x+f(y))-2((c+1)x+f(y)) \geq 2(f(y)-2y),$

implying that

$(f(x+2y)+2cx)-2((c+1)x+f(y)) \geq 2(f(y)-2y)$. Therefore,

$f(x+2y)-2x-2f(y) \geq 2f(y)-4y,$

or equivalently

$f(x+2y) \geq 2x+4f(y)-4y$.

This readily implies that $f(x+2y)-2(x+2y) \geq 4(f(y)-2y)$, and so

$f(x)-2x \geq 4(f(y)-2y)$ for all $x,y>0$ such that $x>2y$. We may generalize this property:

Claim 3: If $x,y>0$ such that $x>2^ny$ for some positive integer $n$, then $f(x)-2x \geq 4^n(f(y)-2y)$.
Proof: We proceed by induction on $n$. The base case is already established. Now, suppose that $f(x)-2x \geq 4^n(f(y)-2y)$ for all $x,y$ such that $x>2^ny$.

Then, if $x>2^{n+1}y$ we may pick an $s$ such that $s \in (2y, \dfrac{x}{2^n})$. Therefore, $x>2^ns$ and $s>2y$. Thus, using the inductive hypothesis and the base case,

$f(x)-2x \geq 4^n(f(s)-2s) \geq 4^{n+1}(f(y)-2y),$

as desired $\blacksquare$

To the problem, let $x,y>0$ such that $x>2y$. Let $n=\lfloor \log_2{\dfrac{x}{y}} \rfloor \geq 1$. Then, $x>2^n y$, and so by Claim 3 we infer that

$f(x)-2x \geq 4^{n}(f(y)-2y) \geq 4^{\log_2{\dfrac{x}{y}}-1}(f(y)-2y),$

or equivalently that

$f(x)-2x \geq \dfrac{x^2(f(y)-2y)}{4y^2}$.

Now, take a $y_0$ such that $f(y_0) \neq 2y_0,$ and let $\dfrac{f(y_0)-2y_0}{4y_0^2}=A>0$. Then,

$f(x) \geq Ax^2+2x$

for all large $x$. Since $A>0$, we obtain that

$f(x)> 4x+1-c$

for all large $x$.

Therefore,

$(f(y)+c+1)-2(2y+1)=f(y)-4y+c-1 >0,$

if $y$ is large, and so by Claim 2 we obtain that

$f(2y+1)+2c=f(c+1+f(y)) \geq 2(c+1+f(y))+2(f(2y+1)-2(2y+1)),$

or equivalently that

$f(2y+1)+2f(y) \leq 8y+2,$

which in light of Claim 1 implies that equality must hold, and so $f(x)=2x$ for all large $x$, say $x>N$. Now, for any $y$ we may take a positive integer $n$ such that $2^ny>N$, and so for all $x>2^ny$ we obtain that $f(x)=2x$. By Claim 3, this implies that

$0=f(x)-2x \geq 4^n(f(y)-2y) \geq 0,$

and so equality must hold everywhere, thus $f(x)=2x$ for all $x$, which is a contradiction to our initial assumption.

To sum up, the only solution is $f(x)=2x$ for all $x>0$.
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vsamc
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#7
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Solution
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TLP.39
778 posts
#8 • 2 Y
Y by StarLex1, Zaro23
CyclicISLscelesTrapezoid wrote:
Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f \colon \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that \[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\]

$P(x+2z,y)-P(x+2y,z)$ implies that $\frac{(2z(c+1)+f(y))-(2y(c+1)+f(z))}{4c(z-y)}$ is constant (for $y\neq z$). The rest is easy.

(The linked lemma can't be applied directly, but the same proof can be used here.)
This post has been edited 1 time. Last edited by TLP.39, Jul 6, 2023, 3:53 AM
Reason: clarification
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ltf0501
191 posts
#9
Y by
A standard R+ functional equation lol.

Let $P(x, y)$ be the assertion. First, if there exists $y > 0$ with $f(y) < 2y$, there exists $x > 0$ such that $(c+1)x+f(y)=x+2y$. With the original equation, we deduce that $x = 0$ which is a contradiction. Therefore, $\boxed{f(y) \geq 2y, \, \forall y > 0}$. Then we have
\[
f(x + 2y) + 2cx = f((c + 1)x + f(y)) \geq 2((c + 1)x + f(y)) \implies f(x + 2y) \geq 2(x + f(y)).
\]In particular, $\boxed{f(x + 2y) > f(y)}$ for all $x, y > 0$.

Now we will prove the injectivity. Suppose that $f(a) = f(b)$ for some $a > b > 0$. Compare with $P(x, a)$ and $P(x, b)$ we know that $f(2a + x) =f(2b + x)$. It is a periodic function with $f(2b) = f(2b + (2a - 2b) \cdot k)$ for all $k \in \mathbb{N}$. We can choose $k$ sufficiently large so that $2b + (2a - 2b) \cdot k > 2 \cdot 2b$. Combined with the previous inequality, it is a contradiction. Hence $f$ is injective.

Next, compared with $P(\frac{z}{c+1} + 2x, y)$ and $P(\frac{z}{c + 1} + 2y, x)$ we have
\[
f(2(c + 1)x + f(y) + z) - f(2(c + 1)y + f(x) + z) = 4(x - y).
\]Also, compared with $P(2x, y)$ and $P(2y, x)$, the above equation also holds for $z = 0$. For given $x, y > 0$, we select $t > x, y$, then
\[
f(2(c + 1)(x + t) + f(y + t)) - f(2(c + 1)(y + t) + f(x + t)) = 4(x - y).
\]Since $f(x + t) > f(x)$ and $2(c + 1)(y + t) > 2(c + 1)y$, there exists $z > 0$ with
\[2(c + 1)y + f(x) + z = 2(c + 1)(y + t) + f(x + t).\]Compared with the previous two equations and by the injectivity of $f$ we get
\[2(c + 1)x + f(y) + z = 2(c + 1)(x + t) + f(y + t).\]Thus we have
\[
f(x) - f(y) = f(x + t) - f(y + t).
\]The equation also holds for all $t > 0$ since for given $x, y, z > 0$ we can choose a $A > \max\{x + 2z, y + 2z\}$, then
\[
f(x) - f(y) = f(x + A) - f(y + A) = f((x + z) + (A - z)) - f((y + z) + (A - z)) = f(x + z) - f(y + z)
\]
With this strong equation, we can substitute the original equation into
\[
Q(x, y): f(cx + f(y)) = f(2y) + 2cx \implies f(x + f(y)) = f(2y) + 2x
\]Compared with $Q(f(x), y)$ and $Q(f(y), x)$ we have $f(2y) = 2f(y)$, so
\[
Q(x, y) \implies f(x + f(y)) = 2(x + f(y)),
\]which also implies there exists a constant $M$ such that $f(t) = 2t$ for all $t > M$. Finally, for all $x > 0$, there exists $k \in \mathbb{N}$ such that $2^k x > M$, then we have $f(x) = \frac{f(2^kx)}{2^k} = \frac{2 \cdot 2^kx}{2^k} = 2x$. Hence the unique solution is $\boxed{f(x) \equiv 2x}$.
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Prod55
127 posts
#10 • 2 Y
Y by Batapan, PRMOisTheHardestExam
Let $P(x,y)$ be the given assertion

$(c+1)x+f(y)\neq x+2y\Rightarrow 2y-f(y)\leq 0\Leftrightarrow f(x)\geq 2x,x>0$

Define $g:\mathbb{R}_{>0}\to \mathbb{R}_{\geq 0}$, $ g(x)=f(x)-2x$

Now $P(x,y)$ becomes $Q(x,y): 2g(y)+g((x+1)c+f(y))=g(x+2y)$.

$Q(x,y)\Rightarrow g(x)\geq 2g(y)$ whenever $x> 2y$

Now $(x+1)c+f(y)>2y$ so $g((x+1)c+f(y))\geq 2g(y)$ and thus $Q(x,y)\Rightarrow g(x)\geq 4g(y),x> 2y$.

Continuing in the same way we find that $g(x)>kg(y)$, when $x>2y$ for arbitrary large $k$'s and thus $g(y)=0$ etc
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DNCT1
235 posts
#11
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Let $P(x,y)$ denote the assertion. If $f(y)<2y$ for some $y$ then $P\left(x,\frac{2y-f(y)}{c}\right)$ lead the contracdition. So $f(x)\ge 2x\quad\forall x\in\mathbb{R}^+ \ (1)$.
$$P(x,y)\implies f(x+2y)+2cx\ge 2(c+1)x+2f(y)\iff f(x+2y)\ge 2f(y)+2x\quad\forall x,y\in\mathbb{R}^+.$$Let denote this assertion is $Q(x,y)$. There are two cases, Let $V= \{x\in\mathbb{R^+}| \ f(x)=2x\}$.
$\textbf{Case 1:}$ $V  \ \text{is empty}$.
We obviously have $f(x)>2x$ for $x>0$. Pick any $u$ such that $g(u)=f(u)-2u=A>0$.
$$P(x-2u,u)\implies f(cx+x-2uc+A)=f(x)+2cx-4cu,\quad\forall x>2u.$$$$Q\left(2cx-4u,x\right)\implies f(cx+x-2uc+A)=f(x)+2cx-4cu\le \frac{f\left(2x+2cx-4cu\right)}{2},\quad\forall x>2u.$$Let $t=cx+x-2uc$ implies that $f(t+A)\le \frac{f(2t)}{2} \quad\forall t>2u. \ (2)$
If $A=f(u)-2u>2u$ then let $t=A$ leads the contracdition, so $f(u)\le 4u$. We claim that if $f(x)>2x$ then $f(x)\le 4x$.
And so in this case $2x<f(x)\le 4x \ \forall x\in\mathbb{R}^+.$ And so $\lim_{x\to 0^+} f(x)=0$.
We use $(2)$ to have $f(2t)\ge 2f(t+A)> 4t+4A \implies f(t)\ge 2t+4A\quad\forall t>4u.$
Let $y>2u$, since $f(y)>2y>4u$, then by $P(x,y)\implies f(x+2y)+2cx> 2(c+1)x+2f(y)+4A\implies f(x+2y)>2x+2f(y)+4A \ \quad\forall x>0, y>2u. \ (99)$
Obvious, we also have $$4x+8y\ge f(x+2y)>2x+2f(y)+4A\quad\forall x>0, y>2u.$$Let $x\to 0$, then $f(y)\le 4y-2A\quad\forall y>2u.$
By $Q(x,y)$, we have $$4x+8y-2A\ge f(x+2y)>2x+2f(y)+4A\quad\forall x>0, y>2u. \ (100)$$Let $x\to 0$ again, we get that $f(y)\le 4y-3A\quad\forall y>2u$, we can do similarly $(100)$ to obtain that $f(y)\le 4y-4A, \ \forall y>2u$. And so $2x+2f(y) \le f(x+2y)\le 4x+8y-4(f(y)-2y)=4x+16y-4f(y)\quad\forall x>0, y>0. \ (101)$
Let $x\to 0$, we get that $f(y)\le \frac{8}{3}y, \ y>0$.
By $(99)$ we have $$2x+2f(y)+4A<f(x+2y)\le \frac{8}{3}(x+2y)\quad\forall x>0, y>2u.\ (102)$$Let $x\to 0$ to see that $f(y)\le \frac{8}{3}y-2A\quad\forall y>2u $, do similarly $(100)$ we have $f(y)\le\frac{8}{3}y-4A$ forall $y>2u$. From this clause, similarly $(102)$, which have $$2x+2f(y)\le f(x+2y)\le \frac{8}{3}(x+2y)-4(f(y)-2y)\quad\forall x,y>0.$$We can do similarly each step in $(100),(101),(102)$ to see that if $f(x)\le v_nx \ \forall x>0$ then $f(x)\le v_{n+1}x$ which sequence $(v_n)$ is determined by $$\begin{cases} \displaystyle \ v_1=4 \\ \ \displaystyle  v_{n+1}=\frac{v_n+4}{3}. \end{cases}$$It's easy to see that $(v_n)$ is decreasing and $\lim v_n=2$, and so $2x< f \le 2x\quad\forall x>0$, which is asburd.
$\textbf{Case 2:} \ \text{there exists} \ a\in V$
$$Q\left(x,\frac{a-x}{2}\right)\implies 2a=f(a)\ge 2f\left(\frac{a-x}{2}\right)+2x\quad\forall 0<x<a.\implies f\left(\frac{a-x}{2}\right)\le a-x\implies f\left(\frac{a-x}{2}\right)=a-x.\ \forall 0<x<a.$$Which means that $f(x)=2x\quad\forall 0<x<\frac{a}{2}.$
$$P\left(x,\frac{a}{8}\right)\implies f\left((c+1)x+\frac{a}{4}\right)=2cx+f\left(x+\frac{a}{4}\right),\quad\forall x>0.$$Let $x\in\left(0,\frac{a}{4}\right)$ implies that $f\left((c+1)x+\frac{a}{4}\right)=2cx+2x+\frac{a}{2}\quad\forall \ 0<x<\frac{a}{4}$.
Let $y=(c+1)x+\frac{a}{4}$ then $f(y)=2y\quad\forall y\in\left(\frac{a}{4},\frac{ca}{4}+\frac{a}{2}\right) \ (3)$.
Denote the sequence $(x_n)$ $$\begin{cases} \displaystyle \ x_1=\frac{a}{2} \\ \ \displaystyle  x_{n+1}=\left(x_n-\frac{a}{4}\right)(c+1)+\frac{a}{4} \ \end{cases} \quad\forall n\in\mathbb{Z}^+.$$It's easy to see that $x_n=\frac{a}{4}(c+1)^{n-1}+\frac{a}{4}$, hence $\lim x_n=+\infty$, and use induction to see that $f(x)=2x\quad\forall x\in \left(\frac{a}{4},x_n\right)$ (continue the process similarly $(3)$ is not hard), but since $f=2x$ in $\left(0,\frac{a}{2}\right)$ so $f(x)=2x\quad\forall x\in (0,x_n)$.
Thus, the only sol is $$\boxed{f(x)=2x\quad\forall x\in\mathbb{R}^+}.$$
This post has been edited 4 times. Last edited by DNCT1, Jul 22, 2023, 5:43 PM
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DS68
29 posts
#12 • 1 Y
Y by PRMOisTheHardestExam
Let $P(x, y)$ be the assertion above.

$\textbf{Claim: } f(x) \geq 2x$
$\textit{Proof}$ If we assume contrary, we can have $P\left(\frac{2y-f(y)}{c}, y\right)$ to reach a contradiction. $\square$

Now define $g: \mathbb{R^+} \rightarrow \mathbb{R}_{\geq 0}, g(x) := f(x) - 2x$. We get that \[g((c+1)x + g(y)+2y)+2g(y) = g(x+2y)\]and redefine $P(x,y)$ for this equation. Notice immediately we get $2g(y) \le g(x+2y)$. As $g$ is bounded on intervals, by Bolzano-Weirstrass theorem, considering a converging sequence $x_n \rightarrow 0$ such that $y = x_i, x+2y = x_j$ gives us that $g(x_n) \rightarrow 0$. As $2g(y) \le g(x+2y)$, we can easily see that every sequence limiting to $0$ gives $g \rightarrow 0$, i.e $\lim_{x \rightarrow 0} g(x) = 0$. Now taking only $y \rightarrow 0$ gives us that $g(Nx) \leq g(x)$ where $N$ can be values arbitrarily close to $c+1$. This implies $g(Mx) \leq g(x)$ for values $M$ arbitrarily close to $(c+1)^n, \forall n \in \mathbb{N}$. Now then having $\frac{g(z+2x)}{2} \geq g(x) \geq g(Mx)$, letting $z = Mx-2x$ for a sufficiently large $n$ gives $g \equiv 0 \rightarrow f \equiv 2x$.
This post has been edited 6 times. Last edited by DS68, Dec 20, 2023, 1:26 PM
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Untro368
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#13
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Let $P(x,y)$ be the original equation. If $f(y)<2y$ for some $y\in\mathbb{R}^+$, then
$$P\left(\dfrac{2y-f(y)}{c},y\right)\implies 2(2y-f(y))=0\ (\text{contradict}).$$So we have $f(y)\geq 2y$. If $f(a)=f(b)$ for some $a-b>0$. Comparing $P(x,a)$ and $P(x,b)$,
$$f(x+d)=f(x)\quad\forall x\gg 0,$$where $d=2(a-b)$. Take $n$ sufficiently large such that exists $x_0>0$ s.t.
$$(c+1)x_0+f(1)=x_0+2,$$then $P(x_0,1)\implies 2cx_0=0$ (contradict). Hence $f$ is injective. Consider
\[f((c+1)x+f(y))+2cz=f(x+2y)+2c(x+z),\]and apply the original equation on both sides, we have
$$f\left((c+1)z+f\left(\dfrac{(c+1)x+f(y)-z}{2}\right)\right)=f\left((c+1)(x+z)+f\left(y-\dfrac{z}{2}\right)\right),$$for $2y>z$ (this implies $f(y)\geq 2y>z$). By injective, we have
$$f\left(\dfrac{(c+1)x+f(y)-z}{2}\right)=(c+1)x+f\left(y-\dfrac{z}{2}\right).$$Compare above equation with $P\left(\dfrac{(c+1)x}{2c},\dfrac{1}{2}\left(y-\dfrac{z}{2}-\dfrac{(c+1)x}{2}\right)\right)$ and by injective, we have
\[\dfrac{(c+1)x+f(y)-z}{2}=\dfrac{(c+1)^2x}{2c}+f\left(\dfrac{1}{2}\left(y-\dfrac{z}{2}-\dfrac{(c+1)x}{2}\right)\right).\]Let $y=2w+\dfrac{z+(c+1)x}{2}$, we have
$$f\left(2w+\dfrac{z+(c+1)x}{2}\right)=2f(w)+(c+1)x+z \quad \forall x,z>0,$$that is $f(x+2w)=2f(w)+2x$ $\forall w,x\in\mathbb{R}^+$, denoted this equation by $Q(x,w)$. Compare $Q(2x,w)$ and $Q(2w,x)$, we get $f(x)=2x$ for all $x\in\mathbb{R}^+$.
This post has been edited 1 time. Last edited by Untro368, Jul 14, 2023, 9:17 AM
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Strudan_Borisov
16 posts
#14 • 2 Y
Y by PRMOisTheHardestExam, Assassino9931
Let $P(x,y)$ be the given functional equation. Obviously $f(x)=2x$ is a solution. Let's prove that it is the unique one.

Claim 1. $f(x)\geq 2x$.

Proof. Assume there is $x_{0}$ such that $f(x_{0})<2x_{0}$.

$P(\frac{2x_{0}-f(x_{0})}{c}, x_{0})$: $2(2x_{0}-f(x_{0}))=0$ - absurd. The claim follows.

Now, let $g(x)=f(x)-2x$ and rewrite $P(x,y)$ in terms of $g$. We get the following $$g((c+1)x+2y+g(y))+2g(y)=g(x+2y).$$We'll prove that $g \equiv 0$ which will prove the desired result.
Assume there is $y_{0} \in \mathbb{R^{+}}$ such that $g(y_{0})>0$ (the fact that $f(x)\geq 2x$ means that $g(x) \geq 0$). In the last assertion we take $x=p-2y_{0}, y=y_{0}$ to get that: $$g((c+1)p-2y_{0}c+g(y_{0}))+2g(y_{0})=g(p), \forall p>2y_{0}; \longrightarrow Q(p)$$Now, lets fix $p>2y_{0}$ and define the following sequence $\textbf{a} =\{a_{i}\}_{i=0}^{\infty}: a_{0}=p, a_{n}=(c+1)a_{n-1}+g(y_{0})-2cy_{0}$. We see that $a_{n}-a_{n-1}=c(a_{n-1}-2y_{0})+g(y_{0})$. Because of $a_{0}>2y_{0}$ by induction we get that $\textbf{a}$ is strictly increasing, so $a_{n}>2{y_0}$. It means that $g(a_{i+1})+2g(y_{0})=g(a_{i})$ from $Q(a_{i})$. Summing the last one for $i=0,1,...,n-1$ we get that $g(a_{n})+2ng(y_{0})=g(a_{0})$, so $g(a_{0})\geq2ng(y_{0})$ which contradicts the fact $n$ can tend to infinity. So, $g\equiv 0$ and we are done.
This post has been edited 2 times. Last edited by Strudan_Borisov, Jul 30, 2023, 5:57 PM
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AlperenINAN
89 posts
#15 • 2 Y
Y by PRMOisTheHardestExam, egxa
Let $P(x,y)$ be the denotation of the given equation.
Lemma: The function $f$ is injective.
Proof: Assume not. Then there exists $a<b$ such that $f(a)=f(b)$. From comparing $P(x,a)$ and $P(x,b)$ we obtain
$$f(x+2a)=f(x+2b)$$In other words $f(x)=f(x+2(b-a))$ for all $x>2a$. Thus $f$ is periodic on $(2a,\infty)$. Let the smallest period be $T$. From comparing $P(x,y)$ and $P(x+\frac{T}{c+1},y)$ we obtain $f(x+2y)=f(x+\frac{T}{c+1}+2y)+\frac{2cT}{c+1}$ and with induction if $z=x+2y$ then
$$f(z)=f(z+\frac{nT}{c+1})+\frac{2cnT}{c+1}$$for all positive integer $n$. But if $n>\frac{(c+1)f(z)}{2cT}$ we'll get a contradiction since the right-hand side is obviously bigger than the left-hand side.
Thus the conclusion follows.
From compairing $P((c+1)x,(c+1)^2x+f(y))$ and $P((3c+1)x,(c+1)x+2y)$ we get the equation
$$f((5c+3)x+4y)+4c^2x=f((2c^2+5c+3)x+2f(y))..........(1)$$From equation $(1)$, $P(2cx,2y+\frac{3(c+1)}{2}x)$ and injectivity we get $f(2y+\frac{3(c+1)}{2}x)=3(c+1)x+2f(y)$. And with rewriting $\frac{3(c+1)}{2}x$ as $x$ we obtain
$$f(x+2y)=2x+2f(y)$$After writing this equation on the original equation and rewriting $(c+1)x$ as $x$ we get
$$f(x+f(y))=2x+2f(y)=f(x+2y)$$With injectivity, we obtain $f(x)=2x$ for all positive reals $x$ and obviously, this function works on the original equation. Thus the only function that satisfies this equation is $f(x)=2x$ for all positive reals $x$
This post has been edited 2 times. Last edited by AlperenINAN, Dec 22, 2023, 10:20 AM
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Cali.Math
128 posts
#16
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We uploaded our solution https://calimath.org/pdf/APMO2023-4.pdf on youtube https://youtu.be/yRc3q7caBvk.
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math_comb01
662 posts
#18
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Stupid Problem,
The only solution is $f(x)=2x$
Claim 1:$f(x) \geq 2x$
Proof
Now the idea is to define $g(x)=f(x)-2x$, the equation now becomes $g((c+1)x+g(y)+2y)+2g(y)=g(x+2y).$
Claim 2 $g \equiv 0$
Proof
Hence we're done
This post has been edited 1 time. Last edited by math_comb01, Jun 17, 2024, 6:11 PM
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solasky
1566 posts
#19
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kred9 wrote:
In particular, $f(2t) = 4t$, so we have $f(y) = 2y$ for all $y < t$ as well, showing that $f(x) = 2x$ for all $x \in \mathbb{R}_{>0}$.
I don’t see why $f(2t) = 4t$ implies that $f(y) = 2y$ for all $y < t$. Could you explain this to me?
This post has been edited 1 time. Last edited by solasky, Mar 27, 2024, 3:37 AM
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Marius_Avion_De_Vanatoare
55 posts
#20
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Looks like all the solutions have very alike ideas, however everybody explains them differently, I will take a try.
First, denote the assertion by $P(x;y)$, assuming there exists $y$ such that $f(y)<2y$, from $P(\frac{2y-f(y)}{c};y)$ we get a contradiction. Assume there exists some $y_1$ such that $f(y_1) \neq 2y_1$, I will achieve a contradiction. First, apply $f(a) \ge a$ to the $LHS$ to obtain after reducing that $2x+2f(y) \le f(x+2y)$. Now define $g:\mathbb{R_+} \rightarrow \mathbb{R}_{\ge0}$, $g(x)=f(x)-2x$. The given property reduces to $g(x+2y) \ge 2g(y)$, call it $Q(x;y)$.
First, I will prove $g$ is unbounded. It is not hard, as $Q(x;x) \Rightarrow g(3x) \ge 2g(x)$, and iterating for $y_1$, we achieve the desired result.
Now I will prove that for any given constant $k$ there exists another constant $N_k$ such that $g(x) \ge kx,  \forall x \ge N_k$.
For this just consider $Q(\epsilon y;y)$ and iterate to get $g((2+\epsilon)^ny_0) \ge 2^ng(y_0)$. Take a sufficiently large $g(y_0)$. Considering the largest $n$ such that $2^ny_0 <x$ , and taking $\epsilon$ such that $(2+\epsilon)^ny_0=x$, I get of course together with $2^ny_0 \ge \frac{x}{2}$, that for $x \ge 2y_0$, $g(x) \ge \frac{x}{2}g(y_0)$, which concludes the claim.
From here, just pick a large enough $k$ and $x \ge N_k$, and applying the inequality $f(x) \ge (2+k)x$ to the $LHS$ of $P(x;y)$, together with small iterations shows that for all big enough $x$, $f(x)$ is unbounded, which obviously can't be true.
This is a very good example of FE, where you use the most standart $\mathbb{R}_+ \rightarrow \mathbb{R}_+$ techniques.
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lnmfx
2 posts
#21
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Solution
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amogususususus
370 posts
#22 • 1 Y
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I hope there is nothing wrong :maybe: .

The only function is $f(x)=2x \ \forall \ x \in \mathbb{R^+}$. Easy to check that this satisfies, now we'll prove that this is the only one. Let $P(x,y)$ be the assertion. All variables mentioned should be in $\mathbb{R^+}$ unless specified.
Claim 1. $f(x)\ge 2x \ \forall \ x \in \mathbb{R^+}$
Proof. Suppose there exist $k \in \mathbb{R^+}$ such that $f(k)<2k$. Then $P(\frac{2k-f(k)}{c},k)$ gives
$$f\left( \frac{2k-f(k)}{c}+f(k) \right)=f\left( \frac{2k-f(k)}{c}+f(k) \right) +2c\left( \frac{2k-f(k)}{c} \right)$$Clearly this isn't possible since $2c\left( \frac{2k-f(k)}{c} \right) >0$. As desired.
Claim 2. $f$ is injective
Proof. Suppose there exist distinct $a,b \in \mathbb{R^+}$ such that $f(a)=f(b)$, we may assume $a>b$.
Comparing $P(x,a)$ and $P(x,b)$ yields
$$f(x+2a)=f(x+2b) \Leftrightarrow f(x)=f(x+(2a-2b)) \ \forall \ x > 2a$$. By Induction, we have
$$f(x)=f(x+n(2a-2b)) \ \forall \ x > 2a $$for all $n \in \mathbb{N}$. Note that $x+n(2a-2b)$ goes to infinity as $n$ goes to infinity.
Since $f(x)\ge 2x \ \forall \ x \in \mathbb{R^+}$, $f(x)$ is infinity for all $x>2a$. Clearly this is absurd because technically infinity is not a real number. So this is a contradiction. As desired.
Claim 3. $f(x)-2x$ is constant
Proof. From $P\left( x+y,\frac{z}{2} \right)$, we have
$$f\left( (c+1)(x+y)+f\left( \frac{z}{2} \right) \right)=f(x+y+z)+2cx+2cy \ (1)$$. From $P\left( x,\frac{y+z}{2} \right)$, we have
$$f\left( (c+1)x+f\left( \frac{y+z}{2} \right) \right)=f(x+y+z)+2cx \ (2)$$. Plugging $(2)$ to $(1)$ yields
$$f\left( (c+1)(x+y)+f\left( \frac{z}{2} \right) \right)=f\left( (c+1)x+f\left( \frac{y+z}{2} \right) \right)+2cy \ $$. Note that $\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2}>0$. From $P\left( y, \frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2}\right)$, we have
$$f\left( (c+1)y + f\left(\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2} \right)\right)=f\left( (c+1)x+f\left( \frac{y+z}{2} \right) \right) + 2cy$$. Injectivity gives
$$(c+1)(x+y)+f\left( \frac{z}{2} \right)= (c+1)y + f\left(\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2} \right)$$$$\Leftrightarrow (c+1)x+f\left( \frac{z}{2} \right)=f\left(\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2} \right)$$. Change $x$ to $\frac{y}{c+1}$ to get
$$y+f\left( \frac{z}{2} \right)=f\left(\frac{f\left( \frac{y+z}{2} \right)}{2} \right)$$. The RHS is symmetric in $y$ and $z$, so swapping them yields
$$y+f\left( \frac{z}{2} \right)=z+f\left( \frac{y}{2} \right) \Leftrightarrow f\left( \frac{y}{2} \right)-y \ is \ constant$$. Just change $y$ to $2y$ and we are done.
Now, plug $f(x)=2x + c \ \forall \ x \in \mathbb{R^+}$ for a constant $c \in \mathbb{R}$ to $P(x,y)$. Easy to check that $c=0$ by pairing the coefficients. As desired.
This post has been edited 1 time. Last edited by amogususususus, Sep 4, 2024, 12:52 PM
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AblonJ
21 posts
#23
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Why This seems too easy.
First note that $f(x) \geq 2x$ , otherwise $P(\frac{2y-f(y)}{c},y)$ gives a contradiction.
Introduce $g(x)=f(x)-2x$. We see that $g(x) \geq 0$.
Now, from the original equation,
\[
g((c+1)x+g(y)+2y)=g(x+2y)-2g(y)
\]As $g((c+1)x+g(y)+2y) \geq 0$, $g(x+2y) \geq 2g(y)$.
\[
g(x+2y)-2g(y)=g((c+1)x+g(y)+2y) \geq 2g(y)
\]Now, $g(x+2y) \geq 4g(y)$. Note that we can continue this process and have that
\[
g(x+2y) \geq Ng(y)
\]for arbitrarily large $N$, which force $g(y)=0$.
Hence, $f(x)=2x$.
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bo18
38 posts
#24
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I don’t want to write why in injective, but I will just say that follows from periodicity

Here is the solution of me:

Given the function:

\[
f((c+1)x+f(y)) = f(x+2y) + 2cx
\]
Let us denote this statement as \( P(x, y) \).

\[
P(x, (c+1)x+f(y))
\]
\[
f((c+1)x+f(x+2y)+2cx) = f(x+2((c+1)x+f(y)))+2cx
\]
The left-hand side can be rewritten as:

\[
f((c+1)x+f(x+2y)+2cx) = f\left( x \cdot \frac{3c+1}{c+1} \cdot (c+1) + f(x+2y) \right)
\]
\[
= f\left( x \cdot \frac{3c+1}{c+1} + 2(x+2y) \right) + 2c x \cdot \frac{3c+1}{c+1}
\]
Thus:

\[
f\left( x \cdot \frac{3c+1}{c+1} + 2(x+2y) \right) + 2c x \cdot \frac{2c}{c+1}
= f(x+2((c+1)x+f(y)))
\]
Applying \( P\left( x \cdot \frac{2c}{c+1}, \frac{4y+3x}{2} \right) \):

\[
f(2cx+f((4y+3x)/2)) = f\left( x \cdot \frac{2c}{c+1} + 4y+3x \right) + 2c x \cdot \frac{2c}{c+1}
\]
\[
= f\left( x \cdot \frac{3c+1}{c+1} + 2(x+2y) \right) + 2c x \cdot \frac{2c}{c+1}
\]
Since this matches the previous expression, we conclude:

\[
f(2cx+f((4y+3x)/2)) = f(x+2((c+1)x+f(y)))
\]
By injectivity:

\[
2cx + f((4y+3x)/2) = x + 2cx + 2x + 2f(y)
\]
Thus:

\[
f((4y+3x)/2) = 3x + 2f(y)
\]
Denoting this as \( Q(x, y) \).

Applying \( Q(x/3, y) \):

\[
f((4y+x)/2) = x + 2f(y)
\]
Denoting this as \( T(x, y) \).

Let \( f(x) = 2x + g(x) \), where \( f(x) > 2x \).

If \( 2x > f(x) \), then applying \( P\left( \frac{2y - f(y)}{c}, y \right) \):

\[
2c \cdot \frac{2y - f(y)}{c} = 0
\]
\[
\Rightarrow 2y = f(y)
\]
which leads to a contradiction.

Thus, from the properties of \( g(x) \), we obtain:

\[
g\left( \frac{4y+x}{2} \right) + \frac{4y+x}{2}\cdot{2}= x + 2g(y) + 4y
\]
\[
\Rightarrow g\left( \frac{4y+x}{2} \right) = 2g(y)
\]
The right-hand side depends on \( x \), but the left-hand side does not.

Substituting:

\[
x \to 4y \Rightarrow g(4y) = 2g(y)
\]
\[
x \to 2y \Rightarrow g(3y) = 2g(y)
\]
\[
x \to 8y \Rightarrow g(6y) = 2g(y)
\]
\[
g(3y) = g(6y)
\]
If we substitute \( y \to y/3 \), we obtain:

\[
g(y) = g(2y)
\]
\[
g(y) = g(2y) = g(4y) = 2g(y)
\]
Thus:

\[
g(y) = 0
\]
Therefore:

\[
f(x) = 2x + g(x) = 2x
\]
This post has been edited 4 times. Last edited by bo18, Mar 8, 2025, 9:00 PM
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SomeonesPenguin
129 posts
#25
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I think this is new.

Claim: $f$ is injective.

sketch

Fix a positive real $a>0$ and let $b=f(a)$.

$\bullet$ $y\mapsto a\implies f((c+1)x+b)=f(x+2a)+2cx$

$\bullet$ $y\mapsto (c+1)y+b\implies f((c+1)x+f(y+2a)+2cy)=f(x+2(c+1)y+2b)+2cx$

\begin{align*}
    f(x+2(c+1)y+2b)+2cx&=f((c+1)x+f(y+2a)+2cy)\\&=f\left(\left(c+1\right)\left(x+\frac{2cy}
    {c+1}\right)+f(y+2a)\right)\\
    &=f\left(x+\frac{2cy}{c+1}+2y+4a\right)+2cx+\frac{4c^2y}{c+1}\\
\end{align*}
If we let $\frac{2c^2}{c+1}=\alpha$ and $4a-2b=\beta$ (which are fixed) we get \[f(x+\alpha y+\beta)=f(x)+2\alpha y\]Henceforth $y\mapsto \frac{cx}{\alpha}$ and $x\mapsto x+2y$ yield \[f(x+2y+cx+\beta)=f(x+2y)+2cx=f((c+1)x+f(y))\]So injectivity further yields $f(y)=2y+\beta$, but this only holds for sufficiently large $y>>\beta $. Anyway, for $x,y>>\beta$ in the original equation we get $\beta=0$ which implies that $f(a)=2a$ for all positive reals $a$, which is indeed a solution.
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jasperE3
11386 posts
#26
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Let $P(x,y)$ be the assertion $f((c+1)x+f(y))=f(x+2y)+2cx$.

Claim 1: $f(x)\ge2x$
Suppose $f(u)<2u$ for some $u$, then:
$P\left(\frac{2u-f(u)}c,u\right)\Rightarrow2(2u-f(u))=0$
which is a contradiction.

Claim 2: $f$ is injective
Suppose $f(a)=f(b)$ for some $a>b$ and let $p=2a-2b>0$. Then:
$P(x,a)\Rightarrow f((c+1)x+f(a))=f(x+2a)+2cx$
$P(x,b)\Rightarrow f((c+1)x+f(a))=f(x+2b)+2cx$
So $f(x+2a)=f(x+2b)$. Then $f(x+p)=f(x)$ for all sufficiently large $x>2b$, and by induction we get that $f(x+np)=f(x)$ for all $x>2b$ and $n\in\mathbb N$. Then fixing $x>2b$ and taking $n\gg\frac{|f(x)-2x|}{2p}$ sufficiently large we get:
$$f(x)=f(x+np)\ge2x+2np>2x+|f(x)-2x|\ge f(x),$$contradiction. Hence $f$ is injective.

Claim 3: if $f(x+a)=f(x)+b$ for all $x>c$ for some $a,b,c\in\mathbb R^+$ then $b=2a$
Note that $f(x+2a)=f(x+a)+b=f(x)+2b$ (again considering only $x>c$). Then:
$P(x,y+a)\Rightarrow f((c+1)x+f(y)+b)=f(x+2y)+2cx+2b$
So we have:
$$f((c+1)x+f(y)+b)=f((c+1)x+f(y))+2b=f((c+1)x+f(y)+2a)$$and so $b=2a$ by injectivity.

Claim 4: $\boxed{f(x)=2x}$ for all $x\in\mathbb R^+$
$P(x,(c+1)y+f(z))\Rightarrow f((c+1)x+f(y+2z)+2cy)=f(x+2(c+1)y+2f(z))+2cx$
$P\left(x+\frac{2cy}{c+1},y+2z\right)\Rightarrow f((c+1)x+f(y+2z)+2cy)=f\left(x+\frac{2cy}{c+1}+2y+4z\right)+2cx+\frac{4c^2y}{c+1}$
Comparing these, we get:
$$f(x+2(c+1)y+2f(z))=f\left(x+\frac{2cy}{c+1}+2y+4z\right)+\frac{4c^2y}{c+1},$$and taking $x\mapsto x-\frac{2cy}{c+1}-2y-4z$ we have:
$$f\left(x+\frac{2c^2y}{c+1}+2f(z)-4z\right)=f(x)+\frac{4c^2y}{c+1}$$for all $x>\frac{2cy}{c+1}+2y+4z$. By Claim 3 (fixing $y,z$) we get $4f(z)-8z=0$, so $f(x)=2x$ for all $x$ which indeed fits.
This post has been edited 4 times. Last edited by jasperE3, Apr 23, 2025, 6:19 PM
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DeathIsAwe
24 posts
#27
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The answer is $f(x) = 2x$ , the check is trivial.

$P(\frac{2y - f(y)}{c}, y) : 0 = 2cx$ which is clearly a contradiction thus $f(x) \geq 2x$ for all $x \in \mathbb{R}^+$.

Now let us manipulate the original equation:
$$f((c+1)x + f(y)) = f(x + 2y) + 2cx$$$$2cx + 2x + 2f(y) \geq f(x + 2y) + 2cx$$$$2x + 2f(y) \geq f(x + 2y)$$Let $g(x) = f(x) - 2x$
$$2x + 4y + 2g(y) \geq 2x + 4y + g(x + 2y)$$$$2g(y) \geq g(x + 2y)$$Now assume that if there exist $a$ s.t. $g(a) = 0$. Notice that for all $x > 2a$, we have $g(x) = 0$.
Then, if you pick $x, y$ in the original equation such that $g(y) > 0$ and $x >2a$, then you have the LFS $= 2x + 4y + 2cx$ and the RHS is larger than $2x + 4y + 2cx$ which is a contradiction, thus if such $a$ exists, we have $f(x) = 2x$ for all $x \in \mathbb{R}^+$.

If no such $g(a)$ exists, then $f(x) > 2x$ for all $x \in \mathbb{R}^+$. Let $[0, b]$ be some fixed interval, and pick any number $k \in [0, b]$. If $\min(g(k)) = 0$ (i.e. there exists $k$ s.t. $g(k)$ is arbritarily close to 0), then notice that for any $x > 2b$, we have to have $g(x) = 0$ which is a contradiction. Thus for any interval $[0, b]$, we have that the lower bound of $g(x)$ in that range is higher than 0.

If the lower bound of $g(x)$ is $d > 0$ (for all $x \in \mathbb{R}^+$), then let $x + 2y$ be a value such that $g(x + 2y)$ is arbitarily close to d. Then the RHS is arbitarily close to $2x + 4y + 2cx + d$ while the LHS must be bigger than $2x + 4y + 2cx + 2d$ which is a contradiction. Thus the lower bound of $g(x)$ is 0.

Now from the interval claim we proved, we can let $y = 1$ and $x + 2y$ is a value such that $g(x + 2y)$ is arbritarily close to 0. In a similar way as before, this creates a contradiction.

Thus $g(x) = 0$ and $f(x) = 2x$.
This post has been edited 1 time. Last edited by DeathIsAwe, Apr 22, 2025, 10:09 AM
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