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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

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


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


More specifically:

For new threads:


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

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


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

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


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


For answers to already existing threads:


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

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



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


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

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
more incircles and tangents
rmtf1111   84
N 2 minutes ago by cj13609517288
Source: EGMO 2019 P4
Let $ABC$ be a triangle with incentre $I$. The circle through $B$ tangent to $AI$ at $I$ meets side $AB$ again at $P$. The circle through $C$ tangent to $AI$ at $I$ meets side $AC$ again at $Q$. Prove that $PQ$ is tangent to the incircle of $ABC.$
84 replies
rmtf1111
Apr 10, 2019
cj13609517288
2 minutes ago
From simple angle chasing <AEP = <AEC = <ADC = 180°- < ADB = 180°- <AFB = 180°
Techno0-8   0
3 minutes ago
From simple angle chasing
<AEP = <AEC = <ADC = 180°- < ADB = 180°- <AFB = 180° - <AFP ( here < is the angle symbol) quadrilateral AFPE is cyclic and <BHC = 180° - < BAC = <EPF = <BPC
and so quadrilateral BHPC is cyclic as well.
By the power of point, CF • CA = CD • CB and CF • CA = CP • CE and CP • CE = CD • CB, hence BEPD is cyclic.

Let AH intersect the circle inscribed in (BPC) at T. <CPT = <CHT = 180°- <AHC = <ABC = <CPD and the points T, D, P are collinear. Now, by applying Pascal's theorem in the hexagon BHTPCC, the intersection points BH and PC, HT and CC, TP and CB are collinear (here CC is tangent) and the points X, L and D are collinear and we are done.
0 replies
Techno0-8
3 minutes ago
0 replies
primes,exponentials,factorials
skellyrah   5
N 4 minutes ago by aaravdodhia
find all primes p,q such that $$ \frac{p^q+q^p-p-q}{p!-q!} $$is a prime number
5 replies
skellyrah
Yesterday at 6:31 PM
aaravdodhia
4 minutes ago
expansion of polynomial
luutrongphuc   0
19 minutes ago
Consider the sequence $(T_n)$ defined by:
\[
T_0 = 0,\quad T_1 = T_2 = 1,
\]and
\[
T_{n+3} = T_{n+2} + T_{n+1} + T_n \quad \text{for all } n \geq 0.
\]Prove that, for every positive integer $n$, we have:
\[
T_{3n} = a_0 T_0 + a_1 T_1 + a_2 T_2 + \cdots + a_{2n} T_{2n},
\]where, for each $i \in \{0, 1, 2, \ldots, 2n\}$, $a_i$ is the coefficient of $x^i$ in the expansion of the polynomial $(x^2 + x + 1)^n$.
0 replies
luutrongphuc
19 minutes ago
0 replies
Do not try to bash on beautiful geometry
ItzsleepyXD   5
N 21 minutes ago by cj13609517288
Source: Own , Mock Thailand Mathematic Olympiad P9
Let $ABC$be triangle with point $D,E$ and $F$ on $BC,AB,CA$
such that $BE=CF$ and $E,F$ are on the same side of $BC$
Let $M$ be midpoint of segment $BC$ and $N$ be midpoint of segment $EF$
Let $G$ be intersection of $BF$ with $CE$ and $\dfrac{BD}{DC}=\dfrac{AC}{AB}$
Prove that $MN\parallel DG$
5 replies
ItzsleepyXD
Yesterday at 9:30 AM
cj13609517288
21 minutes ago
Arbitrary point on BC and its relation with orthocenter
falantrng   28
N 32 minutes ago by Z4ADies
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
28 replies
falantrng
Apr 27, 2025
Z4ADies
32 minutes ago
Cardinality of sets containing both multiples and non-multiples of 3
tom-nowy   1
N an hour ago by Tkn
Source: Own
Let $n$ be a positive integer, and let $N$ be the set $\{ 1,2, \ldots, n\}$.
Let the sets $X_n$ and $Y_n$ be difined as:
\begin{align*}
X_n = &\left\{ (x_1,x_2,x_3) \in N^3  \mid  x_1+x_2+x_3 \text{ is not divisible by } 3. \right\}, \\
Y_n = &\left\{ (y_1,y_2,y_3) \in  N^3  \mid  \text{Among }  y_1-y_2,\, y_2-y_3,\, y_3-y_1, 
\text{ at least one is} \right. \\ 
& \left. \text{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:divisible by }  3  \text{ and at least one is not divisible by } 3. \right\}. 
\end{align*}Which is larger, $\left| X_n \right|$ or $\left| Y_n \right|$ ?
1 reply
tom-nowy
3 hours ago
Tkn
an hour ago
Constant sum
M4RI0   3
N an hour ago by Rohit-2006
Source: Cono Sur Olympiad, Uruguay 1989, Problem #5
Let $ABCD$ be a square with diagonals $AC$ and $BD$, and $P$ a point in one of the sides of the square. Show that the sum of the distances from P to the diagonals is constant.
3 replies
M4RI0
May 14, 2006
Rohit-2006
an hour ago
Four circles
April   56
N an hour ago by ihategeo_1969
Source: Canada Mathematical Olympiad 2007
Let the incircle of triangle $ ABC$ touch sides $ BC,\, CA$ and $ AB$ at $ D,\, E$ and $ F,$ respectively. Let $ \omega,\,\omega_{1},\,\omega_{2}$ and $ \omega_{3}$ denote the circumcircles of triangle $ ABC,\, AEF,\, BDF$ and $ CDE$ respectively.

Let $ \omega$ and $ \omega_{1}$ intersect at $ A$ and $ P,\,\omega$ and $ \omega_{2}$ intersect at $ B$ and $ Q,\,\omega$ and $ \omega_{3}$ intersect at $ C$ and $ R.$

$ a.$ Prove that $ \omega_{1},\,\omega_{2}$ and $ \omega_{3}$ intersect in a common point.

$ b.$ Show that $ PD,\, QE$ and $ RF$ are concurrent.
56 replies
April
Jul 26, 2007
ihategeo_1969
an hour ago
Function equation
LeDuonggg   0
an hour ago
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$:
\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]
0 replies
LeDuonggg
an hour ago
0 replies
If ab+1 is divisible by A then so is a+b
ravengsd   1
N an hour ago by NO_SQUARES
Source: Romania EGMO TST 2025 Day 2, Problem 4
Find the greatest positive integer $A$ such that, for all positive integers $a$ and $b$, if $A$ divides $ab+1$, then $A$ divides $a+b$.
1 reply
ravengsd
2 hours ago
NO_SQUARES
an hour ago
Coincide
giangtruong13   3
N an hour ago by giangtruong13
Source: Hanoi Specialized School's Math Test (Round 2 - Phase 1)
Let $ABCD$ be a trapezoid inscribed in circle $(O)$, $AD||BC, AD < BC$. Let $P$ is the symmetric point of $A$ across $BC$, $AP$ intersects $BC$ at $K$. Let $M$ is midpoint of $BC$ and $H$ is orthocenter of triangle $ABC$. On $BD$ take a point $F$ so that $AF||HM$. Prove that: $ FK,AC,PD$ coincide
3 replies
giangtruong13
Apr 27, 2025
giangtruong13
an hour ago
Geometry..Pls
Jackson0423   0
an hour ago
In equilateral triangle \( ABC \), let \( AB = 10 \). Point \( D \) lies on segment \( BC \) such that \( BC = 4 \cdot DC \). Let \( O \) and \( I \) be the circumcenter and incenter of triangle \( ABD \), respectively. Let \( O' \) and \( I' \) be the circumcenter and incenter of triangle \( ACD \), respectively. Suppose that lines \( OI \) and \( O'I' \) intersect at point \( X \). Find the length of \( XD \).
0 replies
Jackson0423
an hour ago
0 replies
The number of integers
Fang-jh   17
N an hour ago by MathLuis
Source: ChInese TST 2009 P3
Prove that for any odd prime number $ p,$ the number of positive integer $ n$ satisfying $ p|n! + 1$ is less than or equal to $ cp^\frac{2}{3}.$ where $ c$ is a constant independent of $ p.$
17 replies
Fang-jh
Apr 4, 2009
MathLuis
an hour 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
11281 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
8857 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
2533 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
794 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
478 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
1324 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
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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
1416 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
105 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
388 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
1000 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
344 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
571 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
609 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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