Stay ahead of learning milestones! Enroll in a class over the summer!

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

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

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

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
1999 KJMO sum, square sum, cubic sum
RL_parkgong_0106   1
N 13 minutes ago by JH_K2IMO
Source: 1999 KJMO
Three integers are given. $A$ denotes the sum of the integers, $B$ denotes the sum of the square of the integers and $C$ denotes the sum of cubes of the integers(that is, if the three integers are $x, y, z$, then $A=x+y+z$, $B=x^2+y^2+z^2$, $C=x^3+y^3+z^3$). If $9A \geq B+60$ and $C \geq 360$, find $A, B, C$.
1 reply
RL_parkgong_0106
Jun 30, 2024
JH_K2IMO
13 minutes ago
system of equations
JanHaj   4
N 22 minutes ago by justaguy_69
Source: Kosovo National Olympiad 2025, Grade 7, Problem 2
Find all real numbers $a$ and $b$ that satisfy the system of equations:
$$\begin{cases}
 a &= \frac{2}{a+b} \\
 \\
 b &= \frac{2}{3a-b} \\ 
\end{cases}$$
4 replies
JanHaj
Nov 17, 2024
justaguy_69
22 minutes ago
IMO Shortlist 2012, Algebra 2
lyukhson   26
N an hour ago by ezpotd
Source: IMO Shortlist 2012, Algebra 2
Let $\mathbb{Z}$ and $\mathbb{Q}$ be the sets of integers and rationals respectively.
a) Does there exist a partition of $\mathbb{Z}$ into three non-empty subsets $A,B,C$ such that the sets $A+B, B+C, C+A$ are disjoint?
b) Does there exist a partition of $\mathbb{Q}$ into three non-empty subsets $A,B,C$ such that the sets $A+B, B+C, C+A$ are disjoint?

Here $X+Y$ denotes the set $\{ x+y : x \in X, y \in Y \}$, for $X,Y \subseteq \mathbb{Z}$ and for $X,Y \subseteq \mathbb{Q}$.
26 replies
lyukhson
Jul 29, 2013
ezpotd
an hour ago
Inequality with x+y+z=1.
FrancoGiosefAG   3
N an hour ago by JARP091
Let $x,y,z$ be positive real numbers such that $x+y+z=1$. Show that
\[ \frac{x^2-yz}{x^2+x}+\frac{y^2-zx}{y^2+y}+\frac{z^2-xy}{z^2+z}\leq 0. \]
3 replies
FrancoGiosefAG
Yesterday at 8:36 PM
JARP091
an hour ago
f(2x+y)=f(x+2y)
jasperE3   3
N 5 hours ago by MathIQ.
Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that:
$$f(2x+y)=f(x+2y)$$for all $x,y>0$.
3 replies
jasperE3
Yesterday at 8:58 PM
MathIQ.
5 hours ago
Inequalities
sqing   13
N Yesterday at 8:41 PM by ytChen
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
13 replies
sqing
May 13, 2025
ytChen
Yesterday at 8:41 PM
System of Equations
P162008   1
N Yesterday at 8:31 PM by alexheinis
If $a,b$ and $c$ are real numbers such that

$(a + b)(b + c) = -1$

$(a - b)^2 + (a^2 - b^2)^2 = 85$

$(b - c)^2 + (b^2 - c^2)^2 = 75$

Compute $(a - c)^2 + (a^2 - c^2)^2.$
1 reply
P162008
Monday at 10:48 AM
alexheinis
Yesterday at 8:31 PM
Might be the first equation marathon
steven_zhang123   35
N Yesterday at 7:09 PM by lightsbug
As far as I know, it seems that no one on HSM has organized an equation marathon before. Click to reveal hidden textSo why not give it a try? Click to reveal hidden text Let's start one!
Some basic rules need to be clarified:
$\cdot$ If a problem has not been solved within $5$ days, then others are eligible to post a new probkem.
$\cdot$ Not only simple one-variable equations, but also systems of equations are allowed.
$\cdot$ The difficulty of these equations should be no less than that of typical quadratic one-variable equations. If the problem involves higher degrees or more variables, please ensure that the problem is solvable (i.e., has a definite solution, rather than an approximate one).
$\cdot$ Please indicate the domain of the solution to the equation (e.g., solve in $\mathbb{R}$, solve in $\mathbb{C}$).
Here's an simple yet fun problem, hope you enjoy it :P :
P1
35 replies
steven_zhang123
Jan 20, 2025
lightsbug
Yesterday at 7:09 PM
Maximum number of empty squares
Ecrin_eren   0
Yesterday at 6:35 PM


There are 16 kangaroos on a giant 4×4 chessboard, with exactly one kangaroo on each square. In each round, every kangaroo jumps to a neighboring square (up, down, left, or right — but not diagonally). All kangaroos stay on the board. More than one kangaroo can occupy the same square. What is the maximum number of empty squares that can exist after 100 rounds?



0 replies
Ecrin_eren
Yesterday at 6:35 PM
0 replies
THREE People Meet at the SAME. TIME.
LilKirb   7
N Yesterday at 5:33 PM by hellohi321
Three people arrive at the same place independently, at a random between $8:00$ and $9:00.$ If each person remains there for $20$ minutes, what's the probability that all three people meet each other?

I'm already familiar with the variant where there are only two people, where you Click to reveal hidden text It was an AIME problem from the 90s I believe. However, I don't know how one could visualize this in a Click to reveal hidden text Help on what to do?
7 replies
LilKirb
Monday at 1:06 PM
hellohi321
Yesterday at 5:33 PM
Quite straightforward
steven_zhang123   1
N Yesterday at 3:16 PM by Mathzeus1024
Given that the sequence $\left \{ a_{n} \right \} $ is an arithmetic sequence, $a_{1}=1$, $a_{2}+a_{3}+\dots+a_{10}=144$. Let the general term $b_{n}$ of the sequence $\left \{ b_{n} \right \}$ be $\log_{a}{(1+\frac{1}{a_{n}} )} ( a > 0  \text{and}  a \ne  1)$, and let $S_{n}$ be the sum of the $n$ terms of the sequence $\left \{ b_{n} \right \}$. Compare the size of $S_{n}$ with $\frac{1}{3} \log_{a}{(1+\frac{1}{a_{n}} )} $.
1 reply
steven_zhang123
Jan 11, 2025
Mathzeus1024
Yesterday at 3:16 PM
Inequalities
sqing   0
Yesterday at 2:23 PM
Let $ a,b,c>0. $ Prove that$$a^2+b^2+c^2+abc-k(a+b+c)\geq 3k+2-2(k+1)\sqrt{k+1}$$Where $7\geq k \in N^+.$
$$a^2+b^2+c^2+abc-3(a+b+c)\geq-5$$
0 replies
sqing
Yesterday at 2:23 PM
0 replies
Function and Quadratic equations help help help
Ocean_MathGod   1
N Yesterday at 11:26 AM by Mathzeus1024
Consider this parabola: y = x^2 + (2m + 1)x + m(m - 3) where m is constant and -1 ≤ m ≤ 4. A(-m-1, y1), B(m/2, y2), C(-m, y3) are three different points on the parabola. Now rotate the axis of symmetry of the parabola 90 degrees counterclockwise around the origin O to obtain line a. Draw a line from the vertex P of the parabola perpendicular to line a, meeting at point H.

1) express the vertex of the quadratic equation using an expression with m.
2) If, regardless of the value of m, the parabola and the line y=x−km (where k is a constant) have exactly one point of intersection, find the value of k.

3) (where I'm struggling the most) When 1 < PH ≤ 6, compare the values of y1, y2, and y3.
1 reply
Ocean_MathGod
Aug 26, 2024
Mathzeus1024
Yesterday at 11:26 AM
System of Equations
P162008   1
N Yesterday at 10:30 AM by alexheinis
If $a,b$ and $c$ are complex numbers such that

$\frac{ab}{b + c} + \frac{bc}{c + a} + \frac{ca}{a + b} = -9$

$\frac{ab}{c + a} + \frac{bc}{a + b} + \frac{ca}{b + c} = 10$

Compute $\frac{a}{c + a} + \frac{b}{a + b} + \frac{c}{b + c}.$
1 reply
P162008
Monday at 10:34 AM
alexheinis
Yesterday at 10:30 AM
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $x, y, z \in \mathbb R$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
arcsin1.01
27 posts
#2 • 2 Y
Y by Adventure10, HWenslawski
Hint 1
Hint 2
Big hint (do not open before you give up)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
addictedtomath
108 posts
#3 • 2 Y
Y by Adventure10, HWenslawski
solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11364 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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $x$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $\boxed{f(x)=c, \ c \in \mathbb{R}}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
799 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. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Jndd
1417 posts
#17
Y by
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Markas
150 posts
#18
Y by
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
balllightning37
389 posts
#19
Y by
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AshAuktober
1008 posts
#20
Y by
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. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
eg4334
637 posts
#21
Y by
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 $\boxed{f(x) \equiv c}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
blueprimes
356 posts
#22
Y by
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Marcus_Zhang
980 posts
#23
Y by
Almost a one liner
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
575 posts
#24
Y by
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ilikeminecraft
658 posts
#25
Y by
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.
Z K Y
N Quick Reply
G
H
=
a