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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Geometry
AlexCenteno2007   0
2 minutes ago
Let ABC be an acute triangle and A′
the point diametrically opposite A on the circumcircle of the triangle. Through point A, draw a tangent to the circumcircle of triangle ABC that intersects line BC at point D, and take a point E on segment BC such that AD = ED. Let A′′ be the point on the circumcircle of triangle ABC
(other than A) that lies between the reflection of line AA′
and line AE. Show that lines A′A′′ and BC are parallel.
0 replies
AlexCenteno2007
2 minutes ago
0 replies
A diophantine equation
crazyfehmy   14
N 3 minutes ago by MathIQ.
Source: Turkey Junior National Olympiad 2012 P1
Let $x, y$ be integers and $p$ be a prime for which

\[ x^2-3xy+p^2y^2=12p \]
Find all triples $(x,y,p)$.
14 replies
crazyfehmy
Dec 12, 2012
MathIQ.
3 minutes ago
f(2) = 7, find all integer functions [Taiwan 2014 Quizzes]
v_Enhance   59
N 6 minutes ago by MathIQ.
Find all increasing functions $f$ from the nonnegative integers to the integers satisfying $f(2)=7$ and \[ f(mn) = f(m) + f(n) + f(m)f(n) \] for all nonnegative integers $m$ and $n$.
59 replies
v_Enhance
Jul 18, 2014
MathIQ.
6 minutes ago
f(x^3 + y^3 + z^3) = f(x)^3 + f(y)^3 + f(z)^3
pigfly   15
N 10 minutes ago by MathIQ.
Source: VietNam TST 2005, problem 3
Find all functions $f: \mathbb{Z} \mapsto \mathbb{Z}$ satisfying the condition: $f(x^3 +y^3 +z^3 )=f(x)^3+f(y)^3+f(z)^3.$
15 replies
pigfly
Aug 4, 2004
MathIQ.
10 minutes ago
geometry problem
invt   1
N 14 minutes ago by Diamond-jumper76
In a triangle $ABC$ with $\angle B<\angle C$, denote its incenter and midpoint of $BC$ by $I$, $M$, respectively. Let $C'$ be the reflected point of $C$ wrt $AI$. Let the lines $MC'$ and $CI$ meet at $X$. Suppose that $\angle XAI=\angle XBI=90^{\circ}$. Prove that $\angle C=2\angle B$.
1 reply
invt
Yesterday at 11:59 AM
Diamond-jumper76
14 minutes ago
Gergonne point Harmonic quadrilateral
niwobin   3
N an hour ago by niwobin
Triangle ABC has incircle touching the sides at D, E, F as shown.
AD, BE, CF concurrent at Gergonne point G.
BG and CG cuts the incircle at X and Y, respectively.
AG cuts the incircle at K.
Prove: K, X, D, Y form a harmonic quadrilateral. (KX/KY = DX/DY)
3 replies
niwobin
Yesterday at 8:17 PM
niwobin
an hour ago
Computing functions
BBNoDollar   2
N an hour ago by alinazarboland
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
2 replies
BBNoDollar
3 hours ago
alinazarboland
an hour ago
A "side chase" for juniors
Lukaluce   3
N an hour ago by lksb
Source: 2025 Junior Macedonian Mathematical Olympiad P5
Let $M$ be the midpoint of side $BC$ in $\triangle ABC$, and $P \neq B$ is such that the quadrilateral $ABMP$ is cyclic and the circumcircle of $\triangle BPC$ is tangent to the line $AB$. If $E$ is the second common point of the line $BP$ and the circumcircle of $\triangle ABC$, determine the ratio $BE: BP$.
3 replies
Lukaluce
5 hours ago
lksb
an hour ago
IMO ShortList 1998, number theory problem 1
orl   58
N an hour ago by MihaiT
Source: IMO ShortList 1998, number theory problem 1
Determine all pairs $(x,y)$ of positive integers such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$.
58 replies
orl
Oct 22, 2004
MihaiT
an hour ago
constant ratio and angle
k12byda5h   2
N an hour ago by Diamond-jumper76
Source: DGO 2021, Team stage, Day 2 P1
Let triangle $ABC$ be triangle with orthocenter $H$ and circumcircle $O$. A point $X$ lies on line $BC$. $AH$ intersects the circumcircle of triangle $ABC$ again at $H'$. $AX$ intersects circumcircle of triangle $H'HX$ again at $Y$ and intersects circumcircle of triangle $ABC$ again at $Z$. Let $G$ be the intersection of $BC$ with $H'O$. Let $P$ lies on $AB$ such that $PH'A = 90^\circ - \angle BAC$. Prove that
1. the ratio and the angle between $YH$ and $ZG$ do not depend on the choices of $X$.
2. $\angle PYH = \angle BZG$.

Proposed by: k12byda5h
2 replies
k12byda5h
Dec 27, 2021
Diamond-jumper76
an hour ago
Incircle in an isoscoles triangle
Sadigly   3
N 2 hours ago by Diamond-jumper76
Source: own
Let $ABC$ be an isosceles triangle with $AB=AC$, and let $I$ be its incenter. Incircle touches sides $BC,CA,AB$ at $D,E,F$, respectively. Foot of altitudes from $E,F$ to $BC$ are $X,Y$ , respectively. Rays $XI,YI$ intersect $(ABC)$ at $P,Q$, respectively. Prove that $(PQD)$ touches incircle at $D$.
3 replies
Sadigly
Friday at 9:21 PM
Diamond-jumper76
2 hours ago
Canadian MO 2021 P4
MortemEtInteritum   24
N 2 hours ago by Thapakazi
A function $f$ from the positive integers to the positive integers is called Canadian if it satisfies $$\gcd\left(f(f(x)), f(x+y)\right)=\gcd(x, y)$$for all pairs of positive integers $x$ and $y$.

Find all positive integers $m$ such that $f(m)=m$ for all Canadian functions $f$.
24 replies
MortemEtInteritum
Mar 12, 2021
Thapakazi
2 hours ago
Probably a good lemma
Zavyk09   3
N 2 hours ago by MathLuis
Source: found when solving exercises
Let $ABC$ be a triangle with circumcircle $\omega$. Arbitrary points $E, F$ on $AC, AB$ respectively. Circumcircle $\Omega$ of triangle $AEF$ intersects $\omega$ at $P \ne A$. $BE$ intersects $CF$ at $I$. $PI$ cuts $\Omega$ and $\omega$ at $K, L$ respectively. Construct parallelogram $KFRE$. Prove that $A, R, P$ are collinear.
3 replies
Zavyk09
Today at 12:50 PM
MathLuis
2 hours ago
Bushy and Jumpy and the unhappy walnut reordering
popcorn1   53
N 2 hours ago by lksb
Source: IMO 2021 P5
Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the $k$-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$.

Prove that there exists a value of $k$ such that, on the $k$-th move, Jumpy swaps some walnuts $a$ and $b$ such that $a<k<b$.
53 replies
popcorn1
Jul 20, 2021
lksb
2 hours ago
Functions
Potla   23
N Apr 25, 2025 by Ilikeminecraft
Source: 0
Find all functions $ f: \mathbb{R}\longrightarrow \mathbb{R}$ such that
\[f(x+y)+f(y+z)+f(z+x)\ge 3f(x+2y+3z)\]
for all $x, y, z \in \mathbb R$.
23 replies
Potla
Feb 21, 2009
Ilikeminecraft
Apr 25, 2025
Functions
G H J
G H BBookmark kLocked kLocked NReply
Source: 0
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Potla
1886 posts
#1 • 2 Y
Y by Adventure10, HWenslawski
Find all functions $ f: \mathbb{R}\longrightarrow \mathbb{R}$ such that
\[f(x+y)+f(y+z)+f(z+x)\ge 3f(x+2y+3z)\]
for all $x, y, z \in \mathbb R$.
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arcsin1.01
27 posts
#2 • 2 Y
Y by Adventure10, HWenslawski
Hint 1
Hint 2
Big hint (do not open before you give up)
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addictedtomath
108 posts
#3 • 2 Y
Y by Adventure10, HWenslawski
solution
Z K Y
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popcorn1
1098 posts
#4
Y by
$1 + 2 = 3$ again?!

The answer is $f(x)=c$ for all reals $c$; it's easy to see that these solutions work. Let $P(x,y,z)$ denote the given assertion.

Setting $P(x,0,0)$ gives $3f(x) \leq 2f(x) + f(0)$, so $f(x) \leq f(0)$ for all reals $x$. Setting $P(\frac{x}{2},\frac{x}{2},-\frac{x}{2})$ gives $3f(0) \leq 2f(0) + f(x)$, so $f(0) \leq f(x)$ for all reals $x$. Therefore $f(x) \leq f(0) \leq f(x)$, or $f$ is constant.

Remark. I really like this FE because both steps are motivated: you get the first from just doing stuff and you wonder, ``huh, how do I use the inequality?'' Then the solution is natural.
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MrOreoJuice
594 posts
#5 • 2 Y
Y by Mango247, Mango247
Really cute :)
As usual let $P(x,y,z)$ denote the given assertion.
  • $P(x,0,0) \implies f(x) \le f(0)$ for all $x$.
  • $P\left( \dfrac{-3x}{2} , \dfrac{3x}{2} , \dfrac {-x}{2} \right) \implies 2f(0) \le f(-2x) + f(x)$ which is also $\le 2f(0) \implies f(-2x) + f(x) = 2f(0)$.
  • $P(-x , 2x , -x) \implies f(x) \ge f(0)$.
So $f(x) = c$ for constant $c$.

Edit: wait bruh how did I miss the more natural substitution shown above.
This post has been edited 1 time. Last edited by MrOreoJuice, Sep 3, 2021, 10:21 AM
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jasperE3
11352 posts
#6
Y by
$P(x,0,0)\Rightarrow f(x)\le f(0)$
$P(x,x,-x)\Rightarrow f(2x)\ge f(0)\Rightarrow f(x)\ge f(0)\Rightarrow\boxed{f(x)=c}$ for some $c\in\mathbb R$, which works.
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554183
484 posts
#7
Y by
$P(y,y,-y) \implies f(2y)+f(0)+f(0) \geq 3f(0) \implies f(x) \geq f(0) \forall x$
$P(-y,y,-y) \implies f(0)+f(0)+f(-2y) \geq 3f(-2y) \implies f(0) \geq f(-2y) \implies f(0) \geq f(x) \forall x$
Therefore, $f(x)=c$ where $c$ is a constant for all $x$
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HamstPan38825
8867 posts
#8
Y by
Only constant functions work.

Summing this inequality cyclically, $$f(x+2y+3z) + f(y+2z+3x)+f(z+2x+3y) \leq f(x+y) + f(y+z) + f(z+x).$$On the other hand, substituting $P(2x+y, 2y+z, 2z+x)$, we obtain exactly the reverse inequality. As a result, equality must hold everywhere. Now, setting $y=z=0$, $$3f(x) = f(x) + 2f(0) \implies f(x) = f(0)$$is constant, as needed.
This post has been edited 1 time. Last edited by HamstPan38825, Mar 25, 2023, 2:15 AM
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Mogmog8
1080 posts
#9 • 1 Y
Y by centslordm
We claim the only solution is $f(x)=c$, where $c$ is a real constant. Note that this satisfies the given conditions. Let $P(x,y,z)$ denote the given assertion and $P(x,0,0)$ yields $f(x)\le f(0)$. Then, $P(-z,-z,z)$ yields $f(0)\le f(-2z)$ so $f(x)\ge f(0)$. Hence, $f(x)=f(0)$ so we are done. $\square$
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joshualiu315
2534 posts
#10
Y by
Plugging in $(x,y,z)=(x,0,0)$ to get

\[3f(x) \le 2f(x)+f(0)\]\[\implies f(x) \le f(0)\]
Plugging in $(x,y,z)=(x/2,x/2,-x/2)$ gives

\[3f(0) \le f(x)+2f(0)\]\[\implies f(x) \ge f(0)\]
Thus, $f(x)=f(0)$, so our answer is $\boxed{f(x)=c, \ c \in \mathbb{R}}$.
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shendrew7
796 posts
#11
Y by
Substitute $(2k,0,0)$ and $(k,k-k)$ to find
\[f(2k) \leq f(0), \quad f(2k) \ge f(0).\]
Hence the constant $f(0)$ is the only value $f$ can take, so $\boxed{f(x) = c, \quad c \in \mathbb{R}}$, which evidently works. $\blacksquare$
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kamatadu
480 posts
#12
Y by
$f\equiv c$ for some constant $c$ is the solution.

$P(x,0,0) \implies 3f(x) \le f(x) + f(0) + f(x) \implies f(x) \le f(0)$.

$P(-x,-x,x) \implies 3f(-x + 2(-x) + 3x) \le f(-2x) + f(0) + f(0) \implies 3f(0) \le f(-2x) + 2f(0) \implies f(-2x) \ge f(0)$.
Now changing $-2x \rightarrow x$, we get that $f(x) \ge f(0)$.

Thus $f(x) = f(0)$. :yoda:
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dolphinday
1328 posts
#14
Y by
Letting $y = z = 0$ gives $f(x) \leq f(0)$.Then $z = -x = -y$ gives $3f(0) \leq f(2x) + 2f(0)$ so $f(0) \leq f(2x)$, which is only possible if $f(x) = f(0) = c$.
This post has been edited 1 time. Last edited by dolphinday, Mar 6, 2024, 8:40 PM
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eibc
600 posts
#15
Y by
The answer is $f \equiv c$ for any real constant $c$, which works. Let $P(x, y, z)$ denote the given assertion. From $P(x, 0, 0)$ we have $3f(x) \le 2f(x) + f(0)$, so $f(x) \le f(0)$. From $P(\tfrac{x}{2}, \tfrac{x}{2}, -\tfrac{x}{2})$ we have $3f(0) \le f(x) + 2f(0)$, so $f(x) \ge f(0)$. Hence $f(x) = f(0)$ for all $x$, which implies the solution set.
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pikapika007
298 posts
#16
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The answer is $f(x) = c$ only - clearly this works. Now
\[ P(x, 0, 0) \implies 3f(x) \le 2f(x) + f(0) \implies f(x) \le f(0) \]so
\[ 3f(x+2y+3z) \le f(x+y) + f(y+z) + f(z+x) \le 3f(0). \]Now $P(a, a, -a)$ implies that
\[ 3f(0) \le f(2a) + 2f(0) \le 3f(0) \]so $f(0) \le f(a)$ for all $a$, and we're done.
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Jndd
1417 posts
#17
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The answer is $f(x)=c$ for $c\in \mathbb{R}$, and it is easy to see that this satisfies our inequality.

Plugging in $y=z=0$, we get $3f(x)\leq 2f(x) + f(0)$, which gives $f(x)\leq f(0)$ for all $x$. Then, plugging in $x=y=-z$, we get $3f(0)\leq f(-2z)+2f(0)$, so $f(0)\leq f(-2z)$, giving $f(x)\geq f(0)$ for all $x$. Since $f(x)\geq f(0)\geq f(x)$ for all $x$, we must have $f(x)=f(0)$, as desired.
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Markas
150 posts
#18
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Let y = z = 0. Now we plug that in and we get that $3f(x) \leq f(x) + f(0) + f(x)$ $\Rightarrow$ $f(x) \leq f(0)$. Now let x = -z and y = -z. After we plug that in, we get $3f(0) \leq f(-2z) + f(0) + f(0)$ $\Rightarrow$ $f(0) \leq f(-2z)$. By the two substitutions we did we get $f(x) \leq f(0) \leq f(-2z)$ and we can choose x = -2z $\Rightarrow$ we now have $f(x) \leq f(0) \leq f(x)$ $\Rightarrow$ $f(0) = f(x)$ $\Rightarrow$ f is constant. Now we only need to check this which is obviously true since $3c \leq c + c + c$. We are ready since we proved that $f(x) = c$ is the only solution and it works.
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balllightning37
389 posts
#19
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Neat!

$P(x,0,0)$ implies $3f(x)\leq 2f(x)+f(0)$ or $f(x)\leq f(0)$ for all $x$.

Then, $P(x,x,-x)$ implies $3f(0)\leq f(x)+2f(0)$ or $f(0)\leq f(x)$ for all $x$. motivation

Combining, these two, we get that $f(x)=c$ for some constant $c$, which of course works.
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AshAuktober
1008 posts
#20
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Very kawaii, as expected from a Russian problem!
The only function that works is the constant function. It works as then the inequation becomes $3c \ge 3c$ which is true. We now prove it is the only such function.
Let $P(x, y, z)$ note the given assertion.

Claim 1: $f(x) \le f(0)$ for all $x \in \mathbb{R}$.
Proof: $P(x, 0, 0)$ $$\implies 2f(x) + f(0) \ge 3f(x) \implies f(x) \le f(0)$$as desired. $\square$

Claim 2: $f(x) \ge f(0)$ for all $x \in \mathbb{R}$.
Proof: $P\left(\frac x2, \frac x2, -\frac x2 \right)$ $$\implies 2f(0) + f(x) \ge 3f(0) \implies f(x) \ge f(0),$$as desired. $\square$

Combining the above two claims, $f(x) = f(0)$, and thus $f$ is indeed a constant function. $\blacksquare$
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eg4334
637 posts
#21
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Let $x+y=a, y+z=b, x+z=c$ so it rewrites into $$f(a)+f(b)+f(c) \geq 3f(2b+c)$$. Set $b=0, a=c$ to get $2f(a)+f(0) \geq 3f(a) \implies f(0) \geq f(a)$ And then $b=c=0$ to get $f(a) \geq f(0)$. Therefore $f(a)=f(0)$ so $\boxed{f(x) \equiv c}$
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blueprimes
356 posts
#22
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We claim the answer is $f(x) = c$ for a constant $c$. Obviously it works. Now the assertion $y = z = 0$ yields $f(x) \ge f(0)$ for all $x$, whereas $(x, y, z) \mapsto \left(\dfrac{x}{2}, \dfrac{x}{2}, - \dfrac{x}{2} \right)$ yields $f(x) \le f(0)$ for all $x$. So $f(x) = f(0)$ and $f$ is constant as wanted.
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Marcus_Zhang
980 posts
#23
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Almost a one liner
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Maximilian113
575 posts
#24
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Let $P(x, y, z)$ denote the assertion. Then $$P(-x, -x, x) \implies 3f(0) \leq f(-2x)+2f(0) \implies f(x) \geq f(0) \, \forall x \in \mathbb R.$$Also, $$P(-x, x, -x) \implies 3f(-2x) \leq 2f(0)+f(-2x) \implies f(x) \leq f(0) \, \forall x \in \mathbb R.$$It follows that $f(x)=c$ for some constant $c,$ and this clearly satisfies the assertion.
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Ilikeminecraft
657 posts
#25
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I claim that $f(x) = c$ for some constant $c\in\mathbb R$. This clearly works.

Plug in $x = y = -z$ to get that $f(0) \leq f(2x),$ and then $x = -y = z$ to get that $f(2x) \leq f(0).$ Thus, $f(x)$ is constant.
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