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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.

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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
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 2025 KMO
sqing   1
N a minute ago by sqing
Source: Own
Let $ a,b,c,d  $ be real numbers satisfying $ a+b+c+d=0 $ and $ a^2+b^2+c^2+d^2= 6 .$ Prove that $$ -\frac{3}{4} \leq abcd\leq\frac{9}{4}$$Let $ a,b,c,d  $ be real numbers satisfying $ a+b+c+d=6 $ and $ a^2+b^2+c^2+d^2= 18 .$ Prove that $$ -\frac{9(2\sqrt{3}+3)}{4} \leq abcd\leq\frac{9(2\sqrt{3}-3)}{4}$$
1 reply
sqing
15 minutes ago
sqing
a minute ago
RMO 2024 Q1
SomeonecoolLovesMaths   25
N 16 minutes ago by Adywastaken
Source: RMO 2024 Q1
Let $n>1$ be a positive integer. Call a rearrangement $a_1,a_2, \cdots , a_n$ of $1,2, \cdots , n$ nice if for every $k = 2,3, \cdots , n$, we have that $a_1 + a_2 + \cdots + a_k$ is not divisible by $k$.
(a) If $n>1$ is odd, prove that there is no nice arrangement of $1,2, \cdots , n$.
(b) If $n$ is even, find a nice arrangement of $1,2, \cdots , n$.
25 replies
SomeonecoolLovesMaths
Nov 3, 2024
Adywastaken
16 minutes ago
4 variables
Nguyenhuyen_AG   10
N 21 minutes ago by Butterfly
Let $a,\,b,\,c,\,d$ are non-negative real numbers and $0 \leqslant k \leqslant \frac{2}{\sqrt{3}}.$ Prove that
$$a^2+b^2+c^2+d^2+kabcd \geqslant k+4+(k+2)(a+b+c+d-4).$$hide
10 replies
Nguyenhuyen_AG
Dec 21, 2020
Butterfly
21 minutes ago
2025 KMO Inequality
Jackson0423   3
N 26 minutes ago by sqing
Source: 2025 KMO Round 1 Problem 20

Let \(x_1, x_2, \ldots, x_6\) be real numbers satisfying
\[
x_1 + x_2 + \cdots + x_6 = 6,
\]\[
x_1^2 + x_2^2 + \cdots + x_6^2 = 18.
\]Find the maximum possible value of the product
\[
x_1 x_2 x_3 x_4 x_5 x_6.
\]
3 replies
+1 w
Jackson0423
Yesterday at 4:32 PM
sqing
26 minutes ago
All divisors are one more than a perfect power
Tintarn   5
N 34 minutes ago by Nuran2010
Source: Baltic Way 2024, Problem 16
Determine all composite positive integers $n$ such that, for each positive divisor $d$ of $n$, there are integers $k\geq 0$ and $m\geq 2$ such that $d=k^m+1$.
5 replies
Tintarn
Nov 16, 2024
Nuran2010
34 minutes ago
KJMO 2001 P1
RL_parkgong_0106   1
N an hour ago by JH_K2IMO
Source: KJMO 2001
A right triangle of the following condition is given: the three side lengths are all positive integers and the length of the shortest segment is $141$. For the triangle that has the minimum area while satisfying the condition, find the lengths of the other two sides.
1 reply
RL_parkgong_0106
Jun 29, 2024
JH_K2IMO
an hour ago
Cotangential circels
CountingSimplex   5
N an hour ago by rong2020
Let $ABC$ be a triangle with circumcenter $O$ and let the angle bisector of $\angle{BAC}$ intersect $BC$
at $D$. The point $M$ is such that $\angle{MCB}=90^o$ and $\angle{MAD}=90^o$. Lines $BM$ and $OA$ intersect at
the point $P$. Show that the circle centered at $P$ and passing through $A$ is tangent to segment
$BC$.
5 replies
CountingSimplex
Jun 23, 2020
rong2020
an hour ago
Inspired by old results
sqing   3
N an hour ago by sqing
Source: Own
Let $a,b,c $ be reals such that $a^2+b^2+c^2=3$ .Prove that
$$(1-a)(k-b)(1-c)+abc\ge -k$$Where $ k\geq 1.$
$$(1-a)(1-b)(1-c)+abc\ge -1$$$$(1-a)(1-b)(1-c)-abc\ge -\frac{1}{2}-\sqrt 2$$
3 replies
sqing
Today at 7:36 AM
sqing
an hour ago
Inspired by Philippine 2025
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b,c,d $ be real numbers . Prove that
$$\frac{(a-1)(b-3)(c-3)(d-1)}{  (a^2+3)(b^2+3)(c^2+3)(d^2+3)} \ge -\frac{7+4\sqrt 3}{144}$$$$\frac{(a-1)(b-2)(c-2)(d-1)}{  (a^2+3)(b^2+3)(c^2+3)(d^2+3)} \ge -\frac{11+4\sqrt 7}{432}$$


1 reply
sqing
an hour ago
sqing
an hour ago
divisible by x+y
Pirkuliyev Rovsen   2
N an hour ago by ytChen
It is known that the number $x^4+y^4-x^2y^2$ is divisible by $(x+y)^2$.(where $x,y{\in}N$ )Prove that the number $x^2$ is divisible by $x+y$
2 replies
Pirkuliyev Rovsen
Feb 24, 2025
ytChen
an hour ago
inequality thing
BinariouslyRandom   3
N an hour ago by sqing
Source: Philippine MO 2025 P5
Find the largest real constant $k$ for which the inequality \[ (a^2+3)(b^2+3)(c^2+3)(d^2+3) + k(a-1)(b-1)(c-1)(d-1) \ge 0 \]holds for all real numbers $a$, $b$, $c$, and $d$.

answer
3 replies
BinariouslyRandom
6 hours ago
sqing
an hour ago
Midpoint in a weird configuration
Gimbrint   0
an hour ago
Source: Own
Let $ABC$ be an acute triangle ($AB<BC$) with circumcircle $\omega$. Point $L$ is chosen arbitrarily on arc $AC$, not containing $B$. Points $D$ and $E$ lie on lines $AB$ and $BC$ respectively, such that $BELD$ - parallelogram. Point $P$ is chosen on arc $BC$, not containing $A$, such that $\angle CBP=\angle BDE$. Line $AP$ intersects $EL$ at $X$, and line $CP$ intersects $DL$ at $Y$. Line $XY$ intersects $AB$, $BC$ and $BP$ at points $M$, $N$ and $T$ respectively.

Prove that $TN=TM$.
0 replies
Gimbrint
an hour ago
0 replies
Find all functions $f$: \(\mathbb{R}\) \(\rightarrow\) \(\mathbb{R}\) such : $(f
guramuta   1
N 2 hours ago by jasperE3
Find all functions $f$: \(\mathbb{R}\) \(\rightarrow\) \(\mathbb{R}\) such :
$(f(x-y))^2= (f(x))^2 - 2f(xy) + (f(y))^2$
1 reply
guramuta
3 hours ago
jasperE3
2 hours ago
Quadrangle, nine-point conic, Steiner line
kosmonauten3114   0
2 hours ago
Source: My own
Let $P_1P_2P_3P_4$ be a general quadrangle which does not form an orthocentric system. Let $P$, $I$, $M$, $T$ be the Euler-Poncelet point ($\text{QA-P2}$), isogonal center ($\text{QA-P4}$), midray homothetic center ($\text{QA-P8}$), inscribed square axes crosspoint ($\text{QA-P23}$) of $P_1P_2P_3P_4$, respectively.
Let $H_1$ be the orthocenter of $\triangle{P_2P_3P_4}$, and define $H_2$, $H_3$, $H_4$ cyclically.
Let $A_{ij}=P_iP_j \cap H_iH_j$ ($\{i, j\} \in \{1, 2, 3, 4\}, i<j$).
Let $B_{ij}=P_iP_j \cap H_kH_l$ ($\{i, j, k, l\} \in \{1, 2, 3, 4\}, i<j$).
Then, the 12 points $A_{12}$, $A_{13}$, $A_{14}$, $A_{23}$, $A_{24}$, $A_{34}$, $B_{12}$, $B_{13}$, $B_{14}$, $B_{23}$, $B_{24}$, $B_{34}$ lie on the same conic, here denoted by $\mathcal{C}_1$.
Let $\mathcal{C}_2$ be the nine-point conic of $P_1P_2P_3P_4$.
Suppose that $\mathcal{C}_1$ and $\mathcal{C}_2$ have 4 distinct real intersection points, and let $U$, $V$, $W$ be the intersections, other than $P$, of $\mathcal{C}_1$ and $\mathcal{C}_2$.

Prove that the Steiner line of $P$ with respect to $\triangle{UVW}$ passes through $I$ and $M$, and show that the center of $\mathcal{C}_1$ and the orthocenter of $\triangle{UVW}$ coincide with $T$.
0 replies
kosmonauten3114
2 hours ago
0 replies
Problem 4 (second day)
darij grinberg   93
N Apr 25, 2025 by Ilikeminecraft
Source: IMO 2004 Athens
Let $n \geq 3$ be an integer. Let $t_1$, $t_2$, ..., $t_n$ be positive real numbers such that \[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\] Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.
93 replies
darij grinberg
Jul 13, 2004
Ilikeminecraft
Apr 25, 2025
Problem 4 (second day)
G H J
Source: IMO 2004 Athens
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shendrew7
799 posts
#83
Y by
We first prove the case $n=3$. Assume FTSOC we have $(t_1,t_2,t_3) = (a,b,a+b+k)$, where $k \ge 0$. Then
\begin{align*}
3^2+1 &> (t_1+t_2+t_3)\left(\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3}\right) \\
&\ge 5 + \frac{t_3}{t_1} + \frac{t_3}{t_2} + \frac{t_1}{t_2} + \frac{t_1}{t_3} \\
&= 7 + \frac{c+k}{b} + \frac{b+k}{c} + \frac{b+c}{b+c+k} \\
&\ge 10 + k\left(\frac 1b + \frac 1c - \frac{1}{b+c+k}\right) \\
&\ge 10,
\end{align*}
contradiction. We now show the case $n \ge 4$. Assume again FTSOC $t_1, t_2, t_3$ do not form the sides of a triangle. Then
\begin{align*}
n^2+1 &> \left(t_1 + t_2 + \ldots + t_n\right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \ldots + \frac{1}{t_n} \right) \\
&= 10 + (n-3)^2 + \left(t_1+\ldots+t_3\right)\left(\frac{1}{t_4}+\ldots+\frac{1}{t_n}\right) + \left(t_4+\ldots+t_n\right) \left(\frac{1}{t_1}+\ldots+\frac{1}{t_3}\right) \\
&\ge 10 + (n-3)^2 + 2\sqrt{10(n-3)^2} \\
&= \left(n+\sqrt{10}-3\right)^2 \\
&\ge n^2+1,
\end{align*}
contradiction. $\blacksquare$
Z K Y
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RedFireTruck
4243 posts
#84
Y by
Assume WLOG that $t_1\le t_2\le \dots\le t_n$.

Claim: $a\ge b+c$ for $a,b,c>0$ means that $\frac{b+c}{a}+\frac{a}{b}+\frac{a}{c}\ge 5$.

Proof: Let $a=b+c+d$. Then, we wish to prove that $(1-\frac{d}{b+c+d})+2+\frac{b}{c}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}\ge 5$. By AM-GM, $\frac{b}{c}+\frac{c}{b}\ge 2$ so it suffices to prove that $\frac{d}{b}+\frac{d}{c}\ge \frac{d}{b+c+d}$, which is true because $\frac1b+\frac1c\ge \frac1{b+c}$.

Notice that for $n=3$, we just need to prove that $t_3\ge t_1+t_2$ implies that $(t_1+t_2+t_3)(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3})\ge 10$. Expanding gives $$(t_1+t_2+t_3)(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3})=3+\frac{t_1}{t_2}+\frac{t_2}{t_1}+\frac{t_1+t_2}{t_3}+\frac{t_3}{t_1}+\frac{t_3}{t_2}\ge 3+2+5=10$$as desired.

Assume that for some $n-1\ge 3$, the assertion is true. We will now prove that the assertion is true for $n$.

If there exists $1\le i<j<k\le n-1$ such that $t_i$, $t_j$, $t_k$ don't form the side lengths of a triangle, then $$(t_1+t_2+\dots+t_{n-1})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n-1}})\ge (n-1)^2+1$$and $$t_n(\frac1{t_1}+\frac1{t_2}+\dots+\frac1{t_{n-1}})+(t_1+t_2+\dots+t_{n-1})\frac{1}{t_n}+1\ge 2n-1$$by AM-GM so $$(t_1+t_2+\dots+t_{n})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n}})\ge n^2+1$$as desired.

Otherwise, assume that $t_n\ge t_1+t_2$. By AM-GM and our claim, $$(t_1+t_2+\dots+t_{n})(\frac{1}{t_1}+\frac{1}{t_2}+\dots+\frac{1}{t_{n}})\ge n+2\binom{n}{2}-4+\frac{t_1+t_2}{t_n}+\frac{t_n}{t_1}+\frac{t_n}{t_2}\ge n^2+1$$as desired.

We are done by induction.
Z K Y
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blueprimes
356 posts
#86
Y by
For the sake of contradiction, assume that three of the $t_i$ are equal to $a$, $b$, $a + b + x$ where $a$, $b$ are positive reals and $x$ is nonnegative. By Cauchy-Schwarz, we have
$$(t_1 + t_2 + \dots + t_n) \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) \ge \left(n - 3 + \sqrt{(2a + 2b + x) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{a + b + x} \right)} \right)^2.$$Now
$$\sqrt{(2a + 2b + x) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{a + b + x} \right)} = \sqrt{6 + \frac{2a}{b} + \frac{2b}{a} + x \left( \frac{1}{a} + \frac{1}{b} - \frac{1}{a + b + x} \right) } \ge \sqrt{10}$$by AM-GM. To reach contradiction it is now sufficient to show $(n - 3 + \sqrt{10})^2 \ge n^2 + 1 \iff n \ge 3$ which is true. We are done.
This post has been edited 2 times. Last edited by blueprimes, Jun 24, 2024, 1:53 PM
Z K Y
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dudade
139 posts
#88
Y by
FTSOC, suppose $t_1 > t_2 + t_3$. Therefore, we want to show
\begin{align*}
n^2 + 1 &> \sum_{1 \leq i<j \leq n} \left(\dfrac{t_i}{t_j} + \dfrac{t_j}{t_i}\right) + n \\
&= \left(\left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right)\right) + \sum_{4\leq i<j \leq n} \left(\dfrac{t_i}{t_j} + \dfrac{t_j}{t_i}\right)+n \\
&\geq \left(\left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right)\right) + \left(n^2 - n - 6\right) + n.
\end{align*}Thus, we want
\begin{align*}
7 > \left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right)
\end{align*}if $t_1 > t_2 + t_3$. By AM-GM-HM, note that
\begin{align*}
\left(\dfrac{t_1}{t_2}+\dfrac{t_2}{t_1}\right) + \left(\dfrac{t_2}{t_3}+\dfrac{t_3}{t_2}\right) + \left(\dfrac{t_3}{t_1}+\dfrac{t_1}{t_3}\right) &= \left(\dfrac{t_2}{t_1} + \dfrac{t_3}{t_1}\right) + \left(\dfrac{t_1}{t_2} + \dfrac{t_1}{t_3}\right) + 2 \\
&\geq \left(\dfrac{t_2}{t_1} + \dfrac{t_3}{t_1}\right)+ \dfrac{4}{\left(\dfrac{t_1}{t_2} + \dfrac{t_1}{t_3}\right)} + 2 = k + \tfrac{4}{k} + 2.
\end{align*}Since $k < 1$, then we must have that $k + \tfrac{4}{k} + 2 \leq 7$ which is clearly a contradiction. Thus, for all $i$, $j$, and $k$, there exists a triangle with side lengths $t_i$, $t_j$, and $t_k$, as desired.
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jolynefag
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#89
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Elegant problem which is still being solved by people these days. So beautiful!
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SomeonesPenguin
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#90 • 1 Y
Y by zzSpartan
storage
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numbertheory97
43 posts
#91
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Really nice problem! I think I accidentally ended up proving the stronger version.

Solution. Without loss of generality, suppose to the contrary that $t_1 \geq t_2 + t_3$. Then \[t_1 + t_2 + t_3 = \frac25 t_1 + \frac35 t_1 + (t_2 + t_3) \geq \frac25 t_1 + \frac85(t_2 + t_3)\]and \[\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} \geq \frac{1}{t_1} + \frac{4}{t_2 + t_3}\]by Cauchy-Schwarz, so we have \[(t_1 + t_2 + \dots + t_n)\left(\frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}\right)\]\[ \geq \left(\frac25 t_1 + \frac85(t_2 + t_3) + t_4 + \dots + t_n\right)\left(\frac{1}{t_1} + \frac{4}{t_2 + t_3} + \frac{1}{t_4} + \dots + \frac{1}{t_n}\right)\]\[ \overset{\text{C-S}}{\geq} \left(\sqrt{\frac25} + \sqrt{\frac{32}{5}} + (n - 3)\right)^2 = \left(n + \sqrt{10} - 3\right)^2.\]But $f(x) = \left(x + \sqrt{10} - 3\right)^2 - x^2$ is strictly increasing, so $f(x) \geq f(3) = 1$, a contradiction. $\square$

Remark. (Motivation) Why $2/5$ and $3/5$? I originally tried using $1/2$'s, which was very close but broke slightly for $n = 3$. But if we use arbitrary coefficients $x$ and $1 - x$, it suffices to maximize the function \[f(x) = \sqrt x + 2\sqrt{2 - x},\]which by C-S or similar methods is greatest when $x = 2/5$. This fits perfectly when $n = 3$.
This post has been edited 1 time. Last edited by numbertheory97, Aug 23, 2024, 11:51 PM
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ezpotd
1286 posts
#92
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Sort $t_n$. Assume to the contrary that there exists some tuple $(t_1 \cdots t_n)$ such that the inequality is satisfied and $t_1 + t_2 \le t_n$. We show that if $n \ge 4$ we can form a similar tuple with $n -1$ elements. Of course, just remove $t_3$, the decrease on the left is $2n - 1$, the decrease on the right is $t_3 \frac{1}{t_3} + \sum_{i \neq 3} \frac{t_i}{t_3} + \frac{t_3}{t_i} \ge 2n - 1$ by AM-GM, so the inequality is preserved. Now we can just reduce to the $n = 3$ case, where we MUST have $10 > 3 + \sum_{sym} \frac{t_i}{t_j}$. Since $\frac{t_3}{t_1} + \frac{t_1}{t_3}, \frac{t_3}{t_2} + \frac{t_2}{t_3}$ are increasing for $t_3 > t_1, t_2$, if the inequality holds true for $t_3 > t_1 + t_2$, it must hold for $t_3 = t_1 + t_2$. Now plugging this value of $t_3$ gives $7 > 2(\frac{t_1}{t_2} + \frac{t_2}{t_1}) + 3$, which is NEVER true by AM-GM, so the original inequality can never hold in the conditions described.
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GeorgeRP
134 posts
#93 • 1 Y
Y by Bumfuzzle
We will solve the problem by induction on n.

Base case: n=3. We have that:
$$10>(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \Leftrightarrow 7abc>\sum{a^2b}$$WLOG we can assume $a$ is the largest. Thus we need to show $a<b+c$. If we rewrite as a quadratic function in respect to $a$ (considering $b$ and $c$ as parameters) we have:
$$0>a^2(b+c)+a(b^2+c^2-7bc)+(b^2c+c^2b)=f(a)$$We however know:
$$ b+c>a \Leftrightarrow \begin{cases}
  f(b+c)\geq0 \\
  b+c>\frac{7bc-b^2-c^2}{2(b+c)} 
\end{cases} \Leftrightarrow \begin{cases}
  2(b+c)^3-8(b+c)bc\geq0\\
  3b^2-3bc+3c^2>0 
\end{cases} \text{ which are both true, finishing the base}$$
IH: for n-1

IS: for n. Let $A=t_1+t_2+\cdots+t_{n-1}, B=\frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_{n-1}}$.WLOG we can assume $t_1\leq t_2 \leq \cdots \leq t_n$. We know that $n^2+1>(A+t_n)(B+\frac{1}{t_n}) \Leftrightarrow n^2-Bt_n-\frac{A}{t_n}>AB$. We will first show that $Bt_n+\frac{A}{t_n}\geq2n-2$. This follows from the fact:
$$Bt_n+\frac{A}{t_n}=\sum_{1\leq i \leq n-1}{\frac{t_n}{t_i}+\frac{t_i}{t_n}} \overset{AM-GM}{\geq} 2n-2$$From this it directly follows that $(n-1)^2+1>AB$, implying that $t_i, t_j, t_k$ are the sides of a triangle for $1\leq i<j<k\leq n-1$. We are now left to show that $t_n<t_1+t_2$ from which the desired will follow (because $t_1, t_2$ are the smallest). For simplicity let $t_1=x, t_2=y$. FTSOC let $t_n=x+y+k$ for some $k\geq0$. We know that by Cauchy-Schwarz $AB\geq(n-1)^2$. We will now show that $t_nB+\frac{A}{t_n}\geq 2n-1$.
$$t_nB+\frac{A}{t_n}= \frac{t_n}{t_1}+\frac{t_1}{t_n}+\frac{t_2}{t_n}+\frac{t_n}{t_2}+(\frac{t_3}{t_n}+\cdots+\frac{t_{n-1}}{t_n})+(\frac{t_n}{t_3}+\cdots+\frac{t_n}{t_{n-1}}) \overset{\text{AM-GM}}{\geq} $$$$ \geq \frac{a+b+k}{a}+\frac{a}{a+b+k}+\frac{a+b+k}{b}+\frac{b}{a+b+k}+2n-6 \geq \frac{a+b}{a}+\frac{a}{a+b}+\frac{a+b}{b}+\frac{b}{a+b}+2n-6=$$$$=\frac{(a+b)^2}{ab}+2n-5\overset{\text{AM-GM}}{\geq}2n-1$$From here it follows that:
$$n^2> AB+t_nB+\frac{A}{t_n}\geq n^2 \text{ \Large{ \lightning }} \text{ with which we are done}$$
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N3bula
288 posts
#94
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Expanding gives us:
\[\left(t_1 + t_2 + \dots + t_n\right)
  \left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) = n+\sum_{i, j \in \{1, 2, \dots, n\}, i\neq j} \frac{t_i}{t_j} +\frac{t_j}{t_i}\]Thus we get if for any $i$, $j$ and $k$ that $\sum_{sym}\frac{t_i}{t_j}\geq 7$ we get a contradiction,
thus it suffices to prove that if we have $a+b\leq c$ that $\sum_{sym}\frac{a}{b}\geq 7$, thus I will
prove that that sum is minimised when $a+b=c$, we get that proving that is equivalent to proving
\[\frac{1}{a}+\frac{1}{b}-\frac{a+b}{c(c+k)}\geq 0\]When $a+b=c$ and $k$ is a positive real. Thus the above is equivalent to:
\[\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\geq 0\]\[\frac{b+a}{ab}-\frac{1}{a+b}\]The above is minimised when $a=b$, thus its equivalent to:
\[\frac{2a}{a^2}-\frac{1}{2a}\geq 0\]Which is clearly true.
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eg4334
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#95
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WLOG $t_1 \leq t_2 \leq \dots \leq t_n$, and FTSOC let $t_i + t_j < t_k$. The condition then gives by Cauchy that $$n^2 +1 \geq \left( n -3 + \sqrt{3+\sum \frac{t_i}{t_j} } \right)^2$$where the summation is taken over all ordered pairs in $t_i, t_j, t_k$. Now notice that $n^2 +1 \geq (n - 3 + \sqrt{10})^2$ has no integer solutions for $n \geq 3$, so we just need to prove that $$\frac{t_i}{t_j} + \frac{t_i}{t_k} + \frac{t_j}{t_i} + \frac{t_j}{t_k} + \frac{t_k}{t_i} + \frac{t_k}{t_j} \geq 7 $$$$\frac{t_i}{t_k} + \frac{t_j}{t_k} + \frac{t_k}{t_i} + \frac{t_k}{t_j} \geq5$$by AMGM. Also, $(\frac{1}{t_i} + \frac{1}{t_j})(t_i + t_j) \geq 4$ by Cauchy so this reduces into $$\frac{t_i+t_j}{t_k} + \frac{4t_k}{t_i+t_i} \geq 5$$which is obvious.
This post has been edited 2 times. Last edited by eg4334, Nov 29, 2024, 8:56 PM
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Maximilian113
575 posts
#96
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We show by induction on $n$ that if there exists at least one triple $(t_i, t_j, t_k)$ that are not the sides of a triangle, then $$n^2+1 \leq (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right).$$$\text{Base Case: }$ For $n=3,$ we must show that if $t_1, t_2, t_3$ cannot be the sides of a triangle, then $$10 \leq (t_1+t_2+t_3)\left( \frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3} \right).$$WLOG, assume that $t_1 \leq t_2 \leq t_3,$ then we have that $t_3 > t_1+t_2.$ Also, because our inequality is homogenous, assume that $t_3=1.$ Letting $x=t_1+t_2,$ it suffices to show that $$7 \leq \frac{t_1}{t_2}+\frac{t_2}{t_1}+x+\frac{1}{t_1}+\frac{1}{t_2}.$$But by AM-GM, $\frac{t_1}{t_2}+\frac{t_2}{t_1} \geq 2,$ so it suffices to show that $$x+\frac{1}{t_1}+\frac{1}{t_2} \geq 5.$$But, $x \leq t_3=1 \implies (x-1)(x-4) \geq 0 \implies 5 \leq x+\frac{4}{x} \leq x+\frac{1}{t_1t_2} \leq x+\frac{1}{t_1}+\frac{1}{t_2}$ by $2$ applications of AM-GM at the end. Therefore, our base case is proven.

$\text{Inductive Step: }$ Now, assume that for $n=k-1 \geq 3$ our proposition holds. Consider the sequence $t_1, t_2, \cdots t_k$ of positive real numbers. Then if there is at least one triple, say WLOG $(t_1, t_2, t_3),$ which are not the sides of a triangle then we have by our inductive hypothesis that
\begin{align*}
(t_1+t_2+\cdots+t_k)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_k} \right) &=(t_1+t_2+\cdots+t_{k-1})\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_{k-1}} \right) + \sum_{i=1}^{k-1} \left( \frac{t_i}{t_k}+\frac{t_k}{t_i} \right)+1 \\
&\geq \left((k-1)^2+1\right)+(k-1)\cdot 2+1 \text{ by AM-GM} \\
&= k^2+1.
\end{align*}Therefore, our induction is complete. QED

Now, for the sake of a contradiction assume that there exists $t_i, t_j, t_k$ which are not the sides of a triangle, and $$n^2+1 > (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right).$$But by our induction above, we have that $$n^2+1 \leq (t_1+t_2+\cdots+t_n)\left( \frac{1}{t_1}+\frac{1}{t_2}+\cdots+\frac{1}{t_n} \right),$$a contradiction. Therefore, all $t_i, t_j, t_k$ are the sides of a triangle. QED
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Marcus_Zhang
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#97
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IMO P4
This post has been edited 1 time. Last edited by Marcus_Zhang, Mar 23, 2025, 3:46 AM
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cubres
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#98
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Yapping
Storage - grinding IMO problems
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Ilikeminecraft
658 posts
#99
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Assume there exists a sequence $t_i$ such that $t_1 \geq t_2 + t_3.$ Then, we have that:
\begin{align*}
  & (t_1 + t_2 + \cdots + t_n)\left(\frac1{t_1} + \frac1{t_2} + \cdots + \frac1{t_n}\right) \\
  & \geq (t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) + (t_1 + t_2 + t_3)\left(\frac1{t_4} + \cdots + \frac1{t_n}\right)\\
  & \qquad + (t_4 + \cdots + t_n)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) + (n - 3)^2 \\
  & \geq (n - 3)^2 + 2(3n - 9) + 3 + (t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right)
\end{align*}Thus, all that is left is to prove that $$(t_1 + t_2 + t_3)\left(\frac1{t_1} + \frac1{t_2} + \frac1{t_3}\right) \geq 10$$Expanding, we have that $$3 + \frac{t_1}{t_2} + \frac{t_3}{t_2} + \frac{t_2 + t_3}{t_1} + \frac{t_1}{t_3} + \frac{t_2}{t_3}$$Now, differentiate w.r.t. $t_1.$ We get $f(t_1) = \frac1{t_2} + \frac1{t_3} - (t_2 + t_3)\frac1{t_1^2}.$ We claim that $t_1^2 \geq t_2t_3.$ However, this is clearly true because $t_1^2 \geq t_2^2 + t_3^2 + 2t_2t_3,$ while $t_2^2 + t_3^2 + t_2 t_3 \geq 0.$ Thus, $f(t_1)$ is increasing as $t_1$ increases. Thus, we can assume that $t_1 = t_2 + t_3.$ This clearly implies that $f(t_1) \geq 10.$ Hence, we are done.
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