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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.

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0 replies
jlacosta
May 1, 2025
0 replies
3 var inequality
SunnyEvan   13
N 39 minutes ago by Nguyenhuyen_AG
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
13 replies
+1 w
SunnyEvan
May 17, 2025
Nguyenhuyen_AG
39 minutes ago
trigonometric inequality
MATH1945   13
N an hour ago by sqing
Source: ?
In triangle $ABC$, prove that $$sin^2(A)+sin^2(B)+sin^2(C) \leq \frac{9}{4}$$
13 replies
MATH1945
May 26, 2016
sqing
an hour ago
Iran TST Starter
M11100111001Y1R   2
N an hour ago by sami1618
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
2 replies
M11100111001Y1R
Tuesday at 7:36 AM
sami1618
an hour ago
Twin Prime Diophantine
awesomeming327.   23
N 2 hours ago by HDavisWashu
Source: CMO 2025
Determine all positive integers $a$, $b$, $c$, $p$, where $p$ and $p+2$ are odd primes and
\[2^ap^b=(p+2)^c-1.\]
23 replies
1 viewing
awesomeming327.
Mar 7, 2025
HDavisWashu
2 hours ago
No more topics!
old and easy imo inequality
Valentin Vornicu   216
N May 24, 2025 by alexanderchew
Source: IMO 2000, Problem 2, IMO Shortlist 2000, A1
Let $ a, b, c$ be positive real numbers so that $ abc = 1$. Prove that
\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]
216 replies
Valentin Vornicu
Oct 24, 2005
alexanderchew
May 24, 2025
old and easy imo inequality
G H J
Source: IMO 2000, Problem 2, IMO Shortlist 2000, A1
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Valentin Vornicu
7301 posts
#1 • 36 Y
Y by kk108, Ali00tb, adityaguharoy, datkz311, samrocksnature, mathematicsy, aj1235666, centslordm, Adventure10, jhu08, megarnie, themathboi101, HWenslawski, Mahmood.sy, Chjun.7, RedFlame2112, ImSh95, Lamboreghini, suvamkonar, RedFireTruck, Lilathebee, OronSH, lian_the_noob12, aidan0626, Mango247, Blue_banana4, Asilbek777, Dynamic_Solver, ItsBesi, cubres, and 6 other users
Let $ a, b, c$ be positive real numbers so that $ abc = 1$. Prove that
\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]
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Peter
3615 posts
#2 • 62 Y
Y by GoJensenOrGoHome, Wave-Particle, jt314, Derive_Foiler, JasperL, samuel, me9hanics, kk108, opptoinfinity, TheImaginaryOne, AyunPa0310, Lord_Eldo_Santos, Ali00tb, AlastorMoody, Polynom_Efendi, adityaguharoy, translate, mathleticguyyy, myh2910, A-Thought-Of-God, ilovepizza2020, CaptainLevi16, samrocksnature, AllanTian, lneis1, centslordm, Adventure10, jhu08, megarnie, timgu, HWenslawski, cttg8217, ImSh95, michaelwenquan, peelybonehead, Kobayashi, Lamboreghini, suvamkonar, RedFireTruck, Lilathebee, Siddharth03, mathboy100, maththeory, straight, EpicBird08, aidan0626, ehuseyinyigit, Mango247, kiyoras_2001, MathIQ., Sedro, Dynamic_Solver, Yiyj1, cubres, and 8 other users
Since $abc = 1$, there exist positive real numbers $x$, $y$ and $z$ such that $a = x/y$, $b = y/z$ and $c = z/x$. The inequality then reduces to

\[ \left( \frac{x-y+z}y \right) \left( \frac{y-z+x}z \right) \left( \frac{z-x+y}x \right) \leq 1.  \]

Substitute $p= b + c - a$, $q = c + a - b$ and $r = a + b - c$. The inequality then reduces to \[ 8pqr \le (p+q)(q+r)(r+p)  \] which is obvious from the AM-GM since at most one of $p,q,r$ can be negaitve..
This post has been edited 3 times. Last edited by Peter, Dec 22, 2005, 11:30 PM
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Valentin Vornicu
7301 posts
#3 • 18 Y
Y by samrocksnature, ion_k, Adventure10, megarnie, ImSh95, michaelwenquan, Lamboreghini, suvamkonar, Lilathebee, Seungjun_Lee, Mango247, Alex-131, cubres, and 5 other users
There's no need to quote my post for giving a solution Peter ;)
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Peter
3615 posts
#4 • 9 Y
Y by samrocksnature, Adventure10, megarnie, ImSh95, michaelwenquan, Lilathebee, Mango247, cubres, and 1 other user
I'm sorry, it's an old habit. :(

Maybe you can configure the resource page so that it ignores quotes? (just like it ignores attachements)
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abdurashidjon
119 posts
#5 • 10 Y
Y by Polynom_Efendi, samrocksnature, Adventure10, megarnie, ImSh95, Lilathebee, Mango247, cubres, and 2 other users
From problem every one can take as
$abc\geq (a+b-c)(a-b+c)(-a+b+c)$ so we have to prove this
WLOG we can assume $a\geq b\geq c$ then $a+c>b$ and $a+b>c$
1) $b+c>a$ an others then it ia sides of triangle so take
$a=x+y, b=y+z, c=z+x$ so from it follows. (we will look next step)
2) $a\geq b+c$ . Since $a,b,c$ are positive so
$abc>0\geq (a+b-c)(a-b+c)(-a+b+c)$
Abdurashid
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José
1828 posts
#6 • 9 Y
Y by samrocksnature, ImSh95, Lilathebee, Adventure10, Mango247, cubres, and 3 other users
Peter VDD wrote:
\[8pqr \le (p+q)(q+r)(r+p)\] which is obvious from the AM-GM since at most one of $p,q,r$ can be negaitve..

I don't understand this step. Please, can anyone explain me?
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Peter
3615 posts
#7 • 9 Y
Y by samrocksnature, timgu, ImSh95, Lilathebee, Adventure10, Mango247, cubres, and 2 other users
note that...
This post has been edited 3 times. Last edited by Peter, Aug 2, 2006, 10:12 AM
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Yimin Ge
253 posts
#8 • 10 Y
Y by adilbek, samrocksnature, Adventure10, ImSh95, Lilathebee, ehuseyinyigit, Mango247, cubres, and 2 other users
José wrote:
Peter VDD wrote:
\[8pqr \le (p+q)(q+r)(r+p)\] which is obvious from the AM-GM since at most one of $p,q,r$ can be negaitve..

I don't understand this step. Please, can anyone explain me?
The numbers $(p+q),(q+r),(r+p)$ are positive by definition.

Now obviously at most one of the numbers $p,q,r$ can be negative. If one of the numbers is negative, then it is trivial since the left side would be negative and the right side positive.

Hence we can assume that they are are all nonnegative and then it is an application of AmGm.
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José
1828 posts
#9 • 7 Y
Y by samrocksnature, ImSh95, Lilathebee, Adventure10, Mango247, cubres, and 1 other user
I feel ignorant. How do I pass from AM-GM inequality to the inequality that Peter gave? I understood everything in his proof but this (the passage)
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Peter
3615 posts
#10 • 7 Y
Y by samrocksnature, Adventure10, ImSh95, hyc721323, Lilathebee, cubres, and 1 other user
you can expand it and apply AM-GM with 8 variables
or you can apply $p+q\ge2\sqrt{pq}$ and multiply each factor
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José
1828 posts
#11 • 7 Y
Y by samrocksnature, Adventure10, ImSh95, Lilathebee, Mango247, cubres, and 1 other user
Now yes, thanks!
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me@home
2349 posts
#12 • 7 Y
Y by samrocksnature, ImSh95, Lilathebee, Adventure10, Mango247, cubres, and 1 other user
Peter VDD wrote:
you can expand it and apply AM-GM with 8 variables
or you can apply $p+q\ge2\sqrt{pq}$ and multiply each factor
Nice options... :rotfl:
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qiuxuezhe
14 posts
#13 • 20 Y
Y by Seekrit, dizzy, YadisBeles, samuel, Drekavac, ziyamelikov, samrocksnature, ZHEKSHEN, mathlearner2357, ImSh95, Lilathebee, ismayilzadei1387, ehuseyinyigit, Adventure10, sabkx, Nuran2010, MrDanic, cubres, and 2 other users
I think this is easier______CHN student
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me@home
2349 posts
#14 • 6 Y
Y by samrocksnature, Adventure10, ImSh95, Lilathebee, cubres, and 1 other user
oh wow thats nice also
hehe i was stuck with that inequality for a while...
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Peter
3615 posts
#15 • 8 Y
Y by jt314, samrocksnature, megarnie, ImSh95, Lilathebee, Adventure10, cubres, and 1 other user
In fact, it's not entirely correct. Why would $x+y-z\ge0$ for example? If not, you can't appy AM-GM. There is a solution in that sense, but not as trivial as the one posted there.
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