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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
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
1 viewing
jlacosta
Yesterday at 11:16 PM
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
4 variables with quadrilateral sides 2
mihaig   6
N 2 minutes ago by mihaig
Source: Own
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$\left(a+b+c+d-2\right)^2+8\geq3\left(abc+abd+acd+bcd\right).$$
6 replies
1 viewing
mihaig
Apr 29, 2025
mihaig
2 minutes ago
Hard inequality
ys33   6
N 3 minutes ago by mihaig
Let $a, b, c, d>0$. Prove that
$\sqrt[3]{ab}+ \sqrt[3]{cd} < \sqrt[3]{(a+b+c)(b+c+d)}$.
6 replies
1 viewing
ys33
6 hours ago
mihaig
3 minutes ago
Nice one
imnotgoodatmathsorry   2
N 3 minutes ago by imnotgoodatmathsorry
Source: Own
With $x,y,z >0$.Prove that: $\frac{xy}{4y+4z+x} + \frac{yz}{4z+4x+y} +\frac{zx}{4x+4y+z} \le \frac{x+y+z}{9}$
2 replies
+1 w
imnotgoodatmathsorry
an hour ago
imnotgoodatmathsorry
3 minutes ago
Permutation with Matrices
SomeonecoolLovesMaths   0
10 minutes ago
Consider all $n \times n$ matrix such that $\forall$ $k \leq n$, $( a_{1k}, a_{2k}, \cdots, a_{nk} )$ is a permutation of $(1,2, \cdots, n)$, call such matrices $\textit{rowgood}$. Consider all $n \times n$ matrix such that $\forall$ $k \leq n$, $( a_{k1}, a_{k2}, \cdots, a_{kn} )$ is a permutation of $(1,2, \cdots, n)$, call such matrices $\textit{columngood}$. How many $n \times n$ matrices exist that are both $\textit{rowgood}$ and $\textit{columngood}$?
0 replies
+1 w
SomeonecoolLovesMaths
10 minutes ago
0 replies
Points U,V,F,E are concyclic (GAMO P5)
Aritra12   4
N 11 minutes ago by ihategeo_1969
Source: GAMO day 2 P5
Let $ABC$ be an acute, non-isosceles triangle, $AD,BE,CF$ be its heights and $(c)$ its circumcircle. $FE$ cuts the circumcircle at points $S,T$, with point $F$ being between points $S,E$. In addition, let $P,Q$ be the midpoints of the major and the minor arc $BC$, respectively. Line $DQ$ cuts $(c)$ at $R$. The circumcircles of triangles $RSF,TER,SFP$ and $TEP$ cut again $PR$ at points $X,Y,Z$ and $W$, respectively. Suppose $(\ell)$ is the line passing through the circumcenters of triangles $AXW,AYZ$ and $(\ell_B ),(\ell_C)$ the parallel lines through $B,C$ to $(\ell)$. If $(\ell_B)$ meets $CF$ at $U$ and $(\ell_C )$ meets $BE$ at $V$, then prove that points $U,V,F,E$ are concyclic.

$\textit{Proposed by Orestis Lignos}$
4 replies
Aritra12
Apr 12, 2021
ihategeo_1969
11 minutes ago
Min and Max
giangtruong13   1
N 13 minutes ago by imnotgoodatmathsorry
Source: PTNK-HCM Specialized School's Practical Math Test (Round 1)
Let $a,b \geq 0$ such that: $a^3+b^3=2$.Prove that: $$  \sqrt[3]{4} \geq a^2-ab+b^2 \geq 1$$
1 reply
giangtruong13
24 minutes ago
imnotgoodatmathsorry
13 minutes ago
Functional Equation f(x(1+y)) = f(x)(1+f(y))
minimario   4
N 23 minutes ago by HuongToiVMO
Solve in $\mathbb{R}: f(x(1+y)) = f(x)(1+f(y))$
4 replies
minimario
Aug 16, 2015
HuongToiVMO
23 minutes ago
Mmo 9-10 graders P5
Bet667   6
N 31 minutes ago by sqing
Let $a,b,c,d$ be real numbers less than 2.Then prove that $\frac{a^3}{b^2+4}+\frac{b^3}{c^2+4}+\frac{c^3}{d^2+4}+\frac{d^3}{a^2+4}\le4$
6 replies
Bet667
Apr 3, 2025
sqing
31 minutes ago
Functional equations in IMO TST
sheripqr   49
N an hour ago by clarkculus
Source: Iran TST 1996
Find all functions $f: \mathbb R \to \mathbb R$ such that $$ f(f(x)+y)=f(x^2-y)+4f(x)y $$ for all $x,y \in \mathbb R$
49 replies
sheripqr
Sep 14, 2015
clarkculus
an hour ago
Good Permutations in Modulo n
swynca   8
N an hour ago by Thapakazi
Source: BMO 2025 P1
An integer $n > 1$ is called $\emph{good}$ if there exists a permutation $a_1, a_2, a_3, \dots, a_n$ of the numbers $1, 2, 3, \dots, n$, such that:
$(i)$ $a_i$ and $a_{i+1}$ have different parities for every $1 \leq i \leq n-1$;
$(ii)$ the sum $a_1 + a_2 + \cdots + a_k$ is a quadratic residue modulo $n$ for every $1 \leq k \leq n$.
Prove that there exist infinitely many good numbers, as well as infinitely many positive integers which are not good.
8 replies
swynca
Apr 27, 2025
Thapakazi
an hour ago
Interesting inequalities
sqing   0
an hour ago
Source: Own
Let $ a,b,c>0 $ and $ (a+b)^2 (a+c)^2=16abc. $ Prove that
$$ 2a+b+c\leq \frac{128}{27}$$$$ \frac{9}{2}a+b+c\leq \frac{864}{125}$$$$3a+b+c\leq 24\sqrt{3}-36$$$$5a+b+c\leq \frac{4(8\sqrt{6}-3)}{9}$$
0 replies
sqing
an hour ago
0 replies
Geometry..Pls
Jackson0423   3
N an hour ago by Jackson0423
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 \( XD \).
3 replies
Jackson0423
Yesterday at 2:43 PM
Jackson0423
an hour ago
How many variables?
Lukaluce   1
N an hour ago by a_507_bc
Source: BMO SL 2024 A1
Let $u, v, w$ be positive reals. Prove that there is a cyclic permutation $(x, y, z)$ of $(u, v, w)$ such that the inequality:
$$\frac{a}{xa + yb + zc} + \frac{b}{xb + yc  + za} + \frac{c}{xc + ya + zb} \ge 
\frac{3}{x + y + z}$$holds for all positive real numbers $a,  b$ and $c$.
1 reply
Lukaluce
an hour ago
a_507_bc
an hour ago
find the radius of circumcircle!
jennifreind   0
an hour ago
In $\triangle \rm ABC$, $  \angle \rm B$ is acute, $\rm{\overline{BC}} = 8$, and $\rm{\overline{AC}} = 3\rm{\overline{AB}}$. Let point $\rm D$ be the intersection of the tangent to the circumcircle of $\triangle \rm ABC$ at point $\rm A$ and the perpendicular bisector of segment $\rm{\overline{BC}}$. Given that $\rm{\overline{AD}} = 6$, find the radius of the circumcircle of $\triangle \rm BCD$.
IMAGE
0 replies
jennifreind
an hour ago
0 replies
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
11287 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
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
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
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
Y by
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
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#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
574 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
611 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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