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
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
Inspired by lgx57
sqing   1
N 5 minutes ago by sqing
Source: Own
Let $ a,b>0, a^4+ab+b^4=60 $. Prove that
$$  2\sqrt{15} \leq a^2+ab+b^2 \leq \frac{3(\sqrt{481}-1)}{4}$$$$\frac{\sqrt{481}-1}{4}\leq a^2-ab+b^2  \leq 2\sqrt{15} $$Let $ a,b>0, a^4-ab+b^4=60 $. Prove that
$$ 2\sqrt{15} \leq a^2+ab+b^2 \leq \frac{3(\sqrt{481}+1)}{4}$$$$ 5<a^2-ab+b^2 \leq 2\sqrt{15} $$
1 reply
2 viewing
sqing
23 minutes ago
sqing
5 minutes ago
Geometry
Lukariman   4
N 30 minutes ago by lbh_qys
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
4 replies
Lukariman
Yesterday at 12:43 PM
lbh_qys
30 minutes ago
Need help with barycentric
Sadigly   0
an hour ago
Hi,is there a good handout/book that explains barycentric,other than EGMO?
0 replies
Sadigly
an hour ago
0 replies
Combinatorics
P162008   3
N an hour ago by P162008
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
3 replies
P162008
4 hours ago
P162008
an hour ago
3-var inequality
sqing   1
N 5 hours ago by sqing
Source: Own
Let $ a,b\geq  0 ,a^3-ab+b^3=1  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{4}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{1}{3}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{1}{3}$$Let $ a,b\geq  0 ,a^3+ab+b^3=3  $. Prove that
$$  \frac{1}{2}\geq     \frac{a}{a^2+3 }+ \frac{b}{b^2+3}   \geq  \frac{1}{4}(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{3}-1)$$$$  \frac{1}{2}\geq     \frac{a}{a^3+3 }+ \frac{b}{b^3+3}   \geq  \frac{1}{2\sqrt[3]{9}}$$$$  \frac{1}{2}\geq \frac{a}{a^2+ab+2}+ \frac{b}{b^2+ ab+2}  \geq  \frac{4\sqrt[3]{3}+3\sqrt[3]{9}-6}{17}$$$$  \frac{1}{2}\geq \frac{a}{a^3+ab+2}+ \frac{b}{b^3+ ab+2}  \geq  \frac{\sqrt[3]{3}}{5}$$
1 reply
sqing
5 hours ago
sqing
5 hours ago
Interesting inequalities
sqing   1
N 6 hours ago by sqing
Source: Own
Let $ a,b\geq  0 ,a^2-ab+b^2+a+b=3  $. Prove that
$$  \frac{39+\sqrt{13}}{78}\geq  \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq  \frac{1}{2}$$Let $ a,b\geq  0 ,a^2+ab+b^2+a+b=3  $. Prove that
$$  \frac{19+\sqrt{10}}{39}\geq  \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq    \frac{39+\sqrt{13}}{78}$$Let $ a,b\geq  0 ,a^2+ab+b^2+a+b=5  $. Prove that
$$  \frac{3}{5}> \frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \geq     \frac{185+3\sqrt{21}}{402}$$
1 reply
sqing
Today at 3:27 AM
sqing
6 hours ago
4-var inequality
sqing   2
N Today at 2:52 AM by sqing
Source: SXTB (4)2025 Q2837
Let $ a,b,c,d> 0  $. Prove that
$$   \frac{1}{(3a+1)^4}+ \frac{1}{(3b+1)^4}+\frac{1}{(3c+1)^4}+\frac{1}{(3d+1)^4} \geq \frac{1}{16(3abcd+1)}$$
2 replies
sqing
Yesterday at 2:59 PM
sqing
Today at 2:52 AM
Inspired by Bet667
sqing   2
N Today at 2:50 AM by sqing
Source: Own
Let $ a,b $ be a real numbers such that $a^2+kab+b^2\ge a^3+b^3.$Prove that$$a+b\leq k+2$$Where $ k\geq 0. $
2 replies
sqing
Yesterday at 2:46 PM
sqing
Today at 2:50 AM
primitive polyominoes
N.T.TUAN   29
N Today at 12:12 AM by Disjunction
Source: USAMO 2007
An animal with $n$ cells is a connected figure consisting of $n$ equal-sized cells[1].

A dinosaur is an animal with at least $2007$ cells. It is said to be primitive it its cells cannot be partitioned into two or more dinosaurs. Find with proof the maximum number of cells in a primitive dinosaur.

(1) Animals are also called polyominoes. They can be defined inductively. Two cells are adjacent if they share a complete edge. A single cell is an animal, and given an animal with $n$ cells, one with $n+1$ cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.
29 replies
N.T.TUAN
Apr 26, 2007
Disjunction
Today at 12:12 AM
Cyclic roots are not real, they can't hurt you
anantmudgal09   21
N Yesterday at 11:49 PM by bjump
Source: INMO 2023 P2
Suppose $a_0,\ldots, a_{100}$ are positive reals. Consider the following polynomial for each $k$ in $\{0,1,\ldots, 100\}$:
$$a_{100+k}x^{100}+100a_{99+k}x^{99}+a_{98+k}x^{98}+a_{97+k}x^{97}+\dots+a_{2+k}x^2+a_{1+k}x+a_k,$$where indices are taken modulo $101$, i.e., $a_{100+i}=a_{i-1}$ for any $i$ in $\{1,2,\dots, 100\}$. Show that it is impossible that each of these $101$ polynomials has all its roots real.

Proposed by Prithwijit De
21 replies
anantmudgal09
Jan 15, 2023
bjump
Yesterday at 11:49 PM
Extremaly hard inequality
blug   1
N Yesterday at 6:25 PM by arqady
Source: Polish Math Olympiad Training Camp 2024
Let $a, b, c$ be non-negative real numbers. Prove that
$$a+b+c+\sqrt{a^2+b^2+c^2-ab-bc-ca}\geq \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+c^2}.$$Looking for an algebraic solution!
1 reply
blug
Yesterday at 5:50 PM
arqady
Yesterday at 6:25 PM
Equation has no integer solution.
Learner94   33
N Yesterday at 5:45 PM by bjump
Source: INMO 2013
Let $a,b,c,d \in \mathbb{N}$ such that $a \ge b \ge c \ge d $. Show that the equation $x^4 - ax^3 - bx^2 - cx -d = 0$ has no integer solution.
33 replies
Learner94
Feb 3, 2013
bjump
Yesterday at 5:45 PM
Problem 1 — Symmetric Squares, Symmetric Products
RockmanEX3   8
N Yesterday at 5:44 PM by Baimukh
Source: 46th Austrian Mathematical Olympiad National Competition Part 1 Problem 1
Let $a$, $b$, $c$, $d$ be positive numbers. Prove that

$$(a^2 + b^2 + c^2 + d^2)^2 \ge (a+b)(b+c)(c+d)(d+a)$$
When does equality hold?

(Georg Anegg)
8 replies
RockmanEX3
Jul 14, 2018
Baimukh
Yesterday at 5:44 PM
Interesting inequality
sealight2107   0
Yesterday at 4:53 PM
Source: Own
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum value of:
$Q=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}$
0 replies
sealight2107
Yesterday at 4:53 PM
0 replies
A three-variable functional inequality on non-negative reals
Tintarn   11
N Apr 12, 2025 by jasperE3
Source: Dutch TST 2024, 1.2
Find all functions $f:\mathbb{R}_{\ge 0} \to \mathbb{R}$ with
\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]for all $x,y,z \in \mathbb{R}_{\ge 0}$.
11 replies
Tintarn
Jun 28, 2024
jasperE3
Apr 12, 2025
A three-variable functional inequality on non-negative reals
G H J
Source: Dutch TST 2024, 1.2
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tintarn
9042 posts
#1 • 3 Y
Y by ehuseyinyigit, Sedro, Parsia--
Find all functions $f:\mathbb{R}_{\ge 0} \to \mathbb{R}$ with
\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]for all $x,y,z \in \mathbb{R}_{\ge 0}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Hamzaachak
61 posts
#2 • 1 Y
Y by JanHaj
The anwser is f(x) = cx^2 (with c is a positif integer or 0)
This post has been edited 1 time. Last edited by Hamzaachak, Jun 29, 2024, 3:25 PM
Reason: Mousse trap
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ACGNmath
327 posts
#3
Y by
We claim that the answer is $\boxed{f(x) = cx^2}$ with $c \geq 0$ a nonnegative real. To see that this works, apply AM-GM:
\[2x^3 z^3 + y^3 = x^3 z^3 + x^3 z^3 + y^3 \geq 3x^2 yz^2.\]We now show that this is the only family of solutions. Let $P(x,y,z)$ denote the proposition $2x^2z f(z) + yf(y) \geq 3yz^2 f(x)$.

$P(0,1,z): f(1) \geq 3z^2 f(0)$. Since $z$ is arbitrary, this means that $f(0)=0$.

$P(x,x,1): 2x^3 f(1) + xf(x) \geq 3xf(x) \quad\Rightarrow\quad 2x^3 f(1) \geq 2xf(x)\quad\Rightarrow\quad f(x) \leq cx^2$
where $c = f(1)$.

$P(1,x,x): 2x f(x) + xf(x) \geq 3cx^3 \quad\Rightarrow\quad f(x) \geq cx^2$.

Combining the two inequalities, we get $f(x) = cx^2$ as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ayeen_izady
32 posts
#4
Y by
Claim: $f(x)=cx^2$ for some $c\ge 0$.
Proof: Let $f(x)=x^2g(x)$. Note that $2x^3z^3g(z)+y^3g(y)\ge 3yx^2z^2g(x)$. By putting $y=xy$ we get that $x^3z^3g(xz)\ge x^3z^3g(x)$. If $x,z>0$ it's clear that $g(x)=c$ for all positive reals $x$ because $g(t)\ge g(s)$ and $g(s)\ge g(t)$ for all positive reals $s,t$. Thus $f(x)=cx^2$ for all positive reals $x$. You can now get that $f(0)=0$ and hence $f(x)=cx^2$
This post has been edited 1 time. Last edited by ayeen_izady, Jun 29, 2024, 2:10 PM
Reason: typo
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Sedro
5845 posts
#5
Y by
ACGNmath wrote:
$P(0,1,z): f(1) \geq 3z^2 f(0)$. Since $z$ is arbitrary, this means that $f(0)=0$.
How do we conclude this? Can't $f(0)$ be negative?
This post has been edited 1 time. Last edited by Sedro, Jun 29, 2024, 2:56 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Hamzaachak
61 posts
#6
Y by
Sedro wrote:
ACGNmath wrote:
$P(0,1,z): f(1) \geq 3z^2 f(0)$. Since $z$ is arbitrary, this means that $f(0)=0$.
How do we conclude this? Can't $f(0)$ be negative?

y=z=0 give f(0) non negatif
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jenishmalla
5 posts
#7
Y by
This was pretty fun and fell into place pretty quick
Motivations for substitution:mostly getting things to add and subtract and making expression less ugly
\[
\text{Let } P(x, y, z) \text{ be the given assertion.}
\]
\[
P(y, y, 1) \Rightarrow y^2 f(1) \geq f(y) \tag{1}
\]\[
P(1, y, y) \Rightarrow f(y) \geq y^2 f(1) \tag{2}
\]
\[
\text{From (1) and (2), we get:} \quad f(y) = y^2 f(1)
\]
\[
\Rightarrow f(x) = kx^2 \quad \text{where } k = f(1)
\]
\[
\text{Now, plug into the original inequality:}
\]\[
k(x^3 z^3 + x^3 z^3 + y^3) \geq k (3x^3 y z^3)
\]
\[
\Rightarrow \text{Inequality holds for all } x, y, z  \text{ if } k \geq 0
\]
\[
\boxed{f(x) = kx^2 \text{ for all } x \text{ where } k \geq 0 \text{ with equality when } x = y = z \text{ or } k = 0}
\]
(sorry for the bad latex)
This post has been edited 2 times. Last edited by jenishmalla, Apr 9, 2025, 4:25 PM
Reason: typo and equality case
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ErTeeEs06
64 posts
#8
Y by
hmm so far there is no completely correct solution in this thread
Sedro is right, f(0) can indeed be any value <=0
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11301 posts
#9
Y by
Isn't it $f(x)=cx^2$ for all $x>0$ and $f(0)\le0$?
adapt something like the solution in #3
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ErTeeEs06
64 posts
#10
Y by
jasperE3 wrote:
Isn't it $f(x)=cx^2$ for all $x>0$ and $f(0)\le0$?
adapt something like the solution in #3

Yeah this is indeed correct.
$f(x)=cx^2$ for all $x>0$ and $f(0)=d$ where $c, d$ are constants with $c\geq 0$ en $d\leq 0$
During TST I was the only one who scored 7 on this cause lot of people didn't see the codomain was $\mathbb{R}$ and not $\mathbb R_{\geq 0}$ and lost a point for that
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Alidq
32 posts
#11 • 1 Y
Y by ErTeeEs06
If we plug $x=0$ and $y=1$ we have $f(1) \geq 3f(0)z^2$, $\forall z\geq 0$. Now if $f(0)>0$ the the RHS is unbounded so we have a contradiction. Thus $f(0) \leq 0$. Substituing $z=0$ we have $yf(y) \geq 0$ so $f(y) \geq 0$ for all $y > 0$ if we take $y=1$ we have $f(1) \geq 0$. Substituting $x=y$ and $z=1$ yields $$2y^3f(1)+yf(y)\geq 3yf(y)$$, so we have $f(y) \leq y^2f(1)$ ,$y>0$ $(1)$. Now if we take $z=y$ and $x=1$ we obtain $$2yf(y)+yf(y) \geq 3y^3f(1)$$so $f(y)\geq y^2f(1)$ , $y>0$ $(2)$.From $(1)$ and $(2)$ we have $f(y)=y^2f(1)$ , $y>0$. Thus $f(x)=kx^2$ where $k>0$ and $f(0)=l$ where $l \leq 0$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jasperE3
11301 posts
#12
Y by
ErTeeEs06 wrote:
jasperE3 wrote:
Isn't it $f(x)=cx^2$ for all $x>0$ and $f(0)\le0$?
adapt something like the solution in #3

Yeah this is indeed correct.
$f(x)=cx^2$ for all $x>0$ and $f(0)=d$ where $c, d$ are constants with $c\geq 0$ en $d\leq 0$
During TST I was the only one who scored 7 on this cause lot of people didn't see the codomain was $\mathbb{R}$ and not $\mathbb R_{\geq 0}$ and lost a point for that

good job
Z K Y
N Quick Reply
G
H
=
a