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non-symmetric ineq (for girls)
easternlatincup   36
N an hour ago by Tony_stark0094
Source: Chinese Girl's MO 2007
For $ a,b,c\geq 0$ with $ a+b+c=1$, prove that

$ \sqrt{a+\frac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\leq \sqrt{3}$
36 replies
easternlatincup
Dec 30, 2007
Tony_stark0094
an hour ago
Divisibility on 101 integers
BR1F1SZ   3
N an hour ago by ClassyPeach
Source: Argentina Cono Sur TST 2024 P2
There are $101$ positive integers $a_1, a_2, \ldots, a_{101}$ such that for every index $i$, with $1 \leqslant i \leqslant 101$, $a_i+1$ is a multiple of $a_{i+1}$. Determine the greatest possible value of the largest of the $101$ numbers.
3 replies
BR1F1SZ
Aug 9, 2024
ClassyPeach
an hour ago
BMO 2021 problem 3
VicKmath7   19
N an hour ago by NuMBeRaToRiC
Source: Balkan MO 2021 P3
Let $a, b$ and $c$ be positive integers satisfying the equation $(a, b) + [a, b]=2021^c$. If $|a-b|$ is a prime number, prove that the number $(a+b)^2+4$ is composite.

Proposed by Serbia
19 replies
VicKmath7
Sep 8, 2021
NuMBeRaToRiC
an hour ago
USAMO 2002 Problem 4
MithsApprentice   89
N 2 hours ago by blueprimes
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 - y^2) = x f(x) - y f(y)  \] for all pairs of real numbers $x$ and $y$.
89 replies
MithsApprentice
Sep 30, 2005
blueprimes
2 hours ago
pqr/uvw convert
Nguyenhuyen_AG   8
N 3 hours ago by Victoria_Discalceata1
Source: https://github.com/nguyenhuyenag/pqr_convert
Hi everyone,
As we know, the pqr/uvw method is a powerful and useful tool for proving inequalities. However, transforming an expression $f(a,b,c)$ into $f(p,q,r)$ or $f(u,v,w)$ can sometimes be quite complex. That's why I’ve written a program to assist with this process.
I hope you’ll find it helpful!

Download: pqr_convert

Screenshot:
IMAGE
IMAGE
8 replies
Nguyenhuyen_AG
Apr 19, 2025
Victoria_Discalceata1
3 hours ago
Inspired by hlminh
sqing   2
N 3 hours ago by SPQ
Source: Own
Let $ a,b,c $ be real numbers such that $ a^2+b^2+c^2=1. $ Prove that $$ |a-kb|+|b-kc|+|c-ka|\leq \sqrt{3k^2+2k+3}$$Where $ k\geq 0 . $
2 replies
sqing
Yesterday at 4:43 AM
SPQ
3 hours ago
A cyclic inequality
KhuongTrang   3
N 3 hours ago by KhuongTrang
Source: own-CRUX
IMAGE
https://cms.math.ca/.../uploads/2025/04/Wholeissue_51_4.pdf
3 replies
KhuongTrang
Monday at 4:18 PM
KhuongTrang
3 hours ago
Tiling rectangle with smaller rectangles.
MarkBcc168   60
N 3 hours ago by cursed_tangent1434
Source: IMO Shortlist 2017 C1
A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even.

Proposed by Jeck Lim, Singapore
60 replies
MarkBcc168
Jul 10, 2018
cursed_tangent1434
3 hours ago
ALGEBRA INEQUALITY
Tony_stark0094   2
N 3 hours ago by Sedro
$a,b,c > 0$ Prove that $$\frac{a^2+bc}{b+c} + \frac{b^2+ac}{a+c} + \frac {c^2 + ab}{a+b} \geq a+b+c$$
2 replies
Tony_stark0094
4 hours ago
Sedro
3 hours ago
Checking a summand property for integers sufficiently large.
DinDean   2
N 3 hours ago by DinDean
For any fixed integer $m\geqslant 2$, prove that there exists a positive integer $f(m)$, such that for any integer $n\geqslant f(m)$, $n$ can be expressed by a sum of positive integers $a_i$'s as
\[n=a_1+a_2+\dots+a_m,\]where $a_1\mid a_2$, $a_2\mid a_3$, $\dots$, $a_{m-1}\mid a_m$ and $1\leqslant a_1<a_2<\dots<a_m$.
2 replies
DinDean
Yesterday at 5:21 PM
DinDean
3 hours ago
Bunnies hopping around in circles
popcorn1   22
N 4 hours ago by awesomeming327.
Source: USA December TST for IMO 2023, Problem 1 and USA TST for EGMO 2023, Problem 1
There are $2022$ equally spaced points on a circular track $\gamma$ of circumference $2022$. The points are labeled $A_1, A_2, \ldots, A_{2022}$ in some order, each label used once. Initially, Bunbun the Bunny begins at $A_1$. She hops along $\gamma$ from $A_1$ to $A_2$, then from $A_2$ to $A_3$, until she reaches $A_{2022}$, after which she hops back to $A_1$. When hopping from $P$ to $Q$, she always hops along the shorter of the two arcs $\widehat{PQ}$ of $\gamma$; if $\overline{PQ}$ is a diameter of $\gamma$, she moves along either semicircle.

Determine the maximal possible sum of the lengths of the $2022$ arcs which Bunbun traveled, over all possible labellings of the $2022$ points.

Kevin Cong
22 replies
1 viewing
popcorn1
Dec 12, 2022
awesomeming327.
4 hours ago
Inequality while on a trip
giangtruong13   14
N Apr 17, 2025 by GeoMorocco
Source: Trip
I find this inequality while i was on a trip, it was pretty fun and i have some new experience:
Let $a,b,c \geq -2$ such that: $a^2+b^2+c^2 \leq 8$. Find the maximum: $$A= \sum_{cyc} \frac{1}{16+a^3}$$
14 replies
giangtruong13
Apr 12, 2025
GeoMorocco
Apr 17, 2025
Inequality while on a trip
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Source: Trip
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giangtruong13
128 posts
#1 • 1 Y
Y by cubres
I find this inequality while i was on a trip, it was pretty fun and i have some new experience:
Let $a,b,c \geq -2$ such that: $a^2+b^2+c^2 \leq 8$. Find the maximum: $$A= \sum_{cyc} \frac{1}{16+a^3}$$
This post has been edited 1 time. Last edited by giangtruong13, Apr 12, 2025, 12:25 PM
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sqing
41782 posts
#2 • 2 Y
Y by cubres, arqady
Let $a,b,c \geq -2$ such that $a^2+b^2+c^2 \leq 8.$ Prove that$$  \frac{1}{16+a^3}+\frac{1}{16+b^3}+\frac{1}{16+c^3}\leq \frac{5}{16}$$pretty fun
This post has been edited 1 time. Last edited by sqing, Apr 12, 2025, 1:48 PM
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giangtruong13
128 posts
#3 • 1 Y
Y by cubres
Bummer tummer
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no_room_for_error
332 posts
#4
Y by
giangtruong13 wrote:
I find this inequality while i was on a trip, it was pretty fun and i have some new experience:
Let $a,b,c \geq -2$ such that: $a^2+b^2+c^2 \leq 8$. Find the maximum: $$A= \sum_{cyc} \frac{1}{16+a^3}$$

$$\frac{1}{16+a^3}=-\frac{a^2(a+2)(a^2-2a+8)}{64(a^3+16)}+\frac{1}{64}a^2+\frac{1}{16}\leq \frac{1}{64}a^2+\frac{1}{16}$$
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GeoMorocco
34 posts
#5
Y by
giangtruong13 wrote:
I find this inequality while i was on a trip, it was pretty fun and i have some new experience:
Let $a,b,c \geq -2$ such that: $a^2+b^2+c^2 \leq 8$. Find the maximum: $$A= \sum_{cyc} \frac{1}{16+a^3}$$
Obviously:
$$\sum_{cyc} \frac{1}{16+a^3} \leq  \sum_{cyc} \frac{1}{16-|a|^3}$$so we only need to study the inequality for negative numbers and the function is convex for $-2 \leq x \leq 0$.

The function is convex with a convex constraint, therefore it is enough to check the boundaries and we get a maximum at $(0,-2,-2)$ and the maximum value is equal to $\frac{5}{16}$.
This post has been edited 2 times. Last edited by GeoMorocco, Apr 16, 2025, 7:53 AM
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arqady
30210 posts
#6
Y by
GeoMorocco wrote:
giangtruong13 wrote:
I find this inequality while i was on a trip, it was pretty fun and i have some new experience:
Let $a,b,c \geq -2$ such that: $a^2+b^2+c^2 \leq 8$. Find the maximum: $$A= \sum_{cyc} \frac{1}{16+a^3}$$

The function is convex...
$$\left( \frac{1}{16+a^3}\right)''=\frac{12a(a^3-8)}{(a^3+16)^3}$$;)
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GeoMorocco
34 posts
#7
Y by
arqady wrote:
$$\left( \frac{1}{16+a^3}\right)''=\frac{12a(a^3-8)}{(a^3+16)^3}$$;)

Obviously:
$$\sum_{cyc} \frac{1}{16+a^3} \leq  \sum_{cyc} \frac{1}{16-|a|^3}$$so we only need to study the inequality for negative numbers and the function is convex for $-2 \leq x \leq 0$.
This post has been edited 1 time. Last edited by GeoMorocco, Apr 16, 2025, 7:52 AM
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arqady
30210 posts
#8
Y by
GeoMorocco wrote:
arqady wrote:
$$\left( \frac{1}{16+a^3}\right)''=\frac{12a(a^3-8)}{(a^3+16)^3}$$;)

Obviously:
$$\sum_{cyc} \frac{1}{16+a^3} \leq  \sum_{cyc} \frac{1}{16-|a|^3}$$so we only need to study the inequality for negative numbers and the function is convex for $-2 \leq x \leq 0$.
Can you write a new conditions and an inequality, that we need to prove now?
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GeoMorocco
34 posts
#9
Y by
arqady wrote:
Can you write a new conditions and an inequality, that we need to prove now?
if at least one of the variables $a,b,c>0$, we get:
$$\sum_{cyc} \frac{1}{16+a^3} < \frac{1}{16+0^3}+ \frac{1}{16+(-2)^3}+\frac{1}{16+(-2)^3}=\frac{1}{16}+\frac{1}{8}+\frac{1}{8}=\frac{5}{16}$$Therefore, it is enough to study the function for $a,b,c\leq 0$. But the function $f(x)=\frac{1}{16+x^3}$ is convex for $-2\leq x\leq 0$ with convex constraints, so it is enough to check the borders where we get a maximum at $(0,-2,-2)$ equal to $\frac{5}{16}$.
This post has been edited 1 time. Last edited by GeoMorocco, Apr 16, 2025, 1:19 PM
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arqady
30210 posts
#10
Y by
GeoMorocco wrote:
arqady wrote:
Can you write a new conditions and an inequality, that we need to prove now?
if at least one of the variables $a,b,c>0$, we get:
$$\sum_{cyc} \frac{1}{16+a^3} < \frac{1}{16+0^3}+ \frac{1}{16+(-2)^3}+\frac{1}{16+(-2)^3}=\frac{1}{16}+\frac{1}{8}+\frac{1}{8}=\frac{5}{16}$$Therefore, it is enough to study the function for $a,b,c\leq 0$. But the function $f(x)=\frac{1}{16+x^3}$ is convex for $-2\leq x\leq 0$ with convex constraints, so it is enough to check the borders where we get a maximum at $(0,-2,-2)$ equal to $\frac{5}{16}$.

For $(-2,-2,-2)$ we obtain a greater value. Why don't we get a greater value with a condition $a^2+b^2+c^2\leq8$?
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GeoMorocco
34 posts
#11
Y by
arqady wrote:
For $(-2,-2,-2)$ we obtain a greater value. Why don't we get a greater value with a condition $a^2+b^2+c^2\leq8$?
You can't choose $(-2,-2,-2)$ as it does not verify the condition $a^2+b^2+c^2 \leq 8$!!!! I know you are a very smart guy :first: , but not sure what do you mean with your suggestion.

Here is the new problem after we proved that it is not optimal to have a positive coordinate:

Let $a,b,c$ be real numbers such as $-2 \leq a,b,c \leq 0$ and $a^2+b^2+c^2 \leq 8$. Prove that:
$$\sum_{cyc} \frac{1}{16+a^3} \leq \frac{5}{16}$$
This post has been edited 2 times. Last edited by GeoMorocco, Apr 17, 2025, 8:24 AM
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arqady
30210 posts
#12 • 1 Y
Y by teomihai
GeoMorocco wrote:
arqady wrote:
For $(-2,-2,-2)$ we obtain a greater value. Why don't we get a greater value with a condition $a^2+b^2+c^2\leq8$?
You can't choose $(-2,-2,-2)$ as it does not verify the condition $a^2+b^2+c^2 \leq 8$!!!!
Read please better my previous post.
The function $f$ is convex on $[-2,0]$ and without condition $a^2+b^2+c^2\leq8$ you can get a maximal value by checking $\{a,b,c\}\subset\{-2,0\}$.
Why with this condition you still can take $\{a,b,c\}\subset\{-2,0\}$? If for any $\{a,b,c\}\subset\{-2,0\}$ you'll obtain that the condition indeed occurs, so your reasoning is true. Otherwise, your reasoning is total wrong, I think.
This post has been edited 1 time. Last edited by arqady, Apr 17, 2025, 8:33 AM
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GeoMorocco
34 posts
#13
Y by
arqady wrote:
Read please better my previous post.
The function $f$ is convex on $[-2,0]$ and without condition $a^2+b^2+c^2\leq8$ you can get a maximal value by checking $\{a,b,c\}\subset\{-2,0\}$.
Why with this condition you still can take $\{a,b,c\}\subset\{-2,0\}$? If for any $\{a,b,c\}\subset\{-2,0\}$ you'll obtain that the condition indeed occurs, so your reasoning is true. Otherwise, your reasoning is total wrong, I think.

I still don't get your point. The condition $a^2+b^2+c^2 \leq 8$ is essential to finding $a,b,c$ not just the border conditions. Check my example:
https://artofproblemsolving.com/community/c6t243f6h3550230_find_the_maxium_of_the_following_expression
This post has been edited 2 times. Last edited by GeoMorocco, Apr 17, 2025, 8:53 AM
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arqady
30210 posts
#14
Y by
GeoMorocco, in your linked problem a maximal value occurs also for $a=-1$ and $-1\notin\{-2,0\}$.
I hope, now you understood me.
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GeoMorocco
34 posts
#15
Y by
arqady wrote:
GeoMorocco, in your linked problem a maximal value occurs also for $a=-1$ and $-1\notin\{-2,0\}$.
I hope, now you understood me.
Yes, of course. if you check my proof earlier, I said: "convex constraints" and not "convex constraint" which means that I used both constraints to find the maximum.
GeoMorocco wrote:
Therefore, it is enough to study the function for $a,b,c\leq 0$. But the function $f(x)=\frac{1}{16+x^3}$ is convex for $-2\leq x\leq 0$ with convex constraints, so it is enough to check the borders where we get a maximum at $(0,-2,-2)$ equal to $\frac{5}{16}$.
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