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JBMO Combinatorics vibes
Sadigly   1
N 44 minutes ago by Royal_mhyasd
Source: Azerbaijan Senior NMO 2018
Numbers $1,2,3...,100$ are written on a board. $A$ and $B$ plays the following game: They take turns choosing a number from the board and deleting them. $A$ starts first. They sum all the deleted numbers. If after a player's turn (after he deletes a number on the board) the sum of the deleted numbers can't be expressed as difference of two perfect squares,then he loses, if not, then the game continues as usual. Which player got a winning strategy?
1 reply
Sadigly
Yesterday at 9:53 PM
Royal_mhyasd
44 minutes ago
line JK of intersection points of 2 lines passes through the midpoint of BC
parmenides51   3
N an hour ago by cursed_tangent1434
Source: Rioplatense Olympiad 2018 level 3 p4
Let $ABC$ be an acute triangle with $AC> AB$. be $\Gamma$ the circumcircle circumscribed to the triangle $ABC$ and $D$ the midpoint of the smallest arc $BC$ of this circle. Let $E$ and $F$ points of the segments $AB$ and $AC$ respectively such that $AE = AF$. Let $P \neq A$ be the second intersection point of the circumcircle circumscribed to $AEF$ with $\Gamma$. Let $G$ and $H$ be the intersections of lines $PE$ and $PF$ with $\Gamma$ other than $P$, respectively. Let $J$ and $K$ be the intersection points of lines $DG$ and $DH$ with lines $AB$ and $AC$ respectively. Show that the $JK$ line passes through the midpoint of $BC$
3 replies
1 viewing
parmenides51
Dec 11, 2018
cursed_tangent1434
an hour ago
Six variables
Nguyenhuyen_AG   2
N an hour ago by arqady
Let $a,\,b,\,c,\,x,\,y,\,z$ be six positive real numbers. Prove that
$$\frac{a}{b+c} \cdot \frac{y+z}{x} + \frac{b}{c+a} \cdot \frac{z+x}{y} + \frac{c}{a+b} \cdot \frac{x+y}{z} \geqslant 2+\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}.$$
2 replies
Nguyenhuyen_AG
Yesterday at 5:09 AM
arqady
an hour ago
Brilliant guessing game on triples
Assassino9931   2
N an hour ago by Mirjalol
Source: Al-Khwarizmi Junior International Olympiad 2025 P8
There are $100$ cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from $1$ to $100$. In a move, Madina is allowed to choose any $3$ cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?

Shubin Yakov, Russia
2 replies
Assassino9931
Saturday at 9:46 AM
Mirjalol
an hour ago
ISI UGB 2025 P5
SomeonecoolLovesMaths   4
N an hour ago by Shiny_zubat
Source: ISI UGB 2025 P5
Let $a,b,c$ be nonzero real numbers such that $a+b+c \neq 0$. Assume that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}$$Show that for any odd integer $k$, $$\frac{1}{a^k} + \frac{1}{b^k} + \frac{1}{c^k} = \frac{1}{a^k+b^k+c^k}.$$
4 replies
SomeonecoolLovesMaths
Yesterday at 11:15 AM
Shiny_zubat
an hour ago
ISI UGB 2025 P2
SomeonecoolLovesMaths   6
N an hour ago by quasar_lord
Source: ISI UGB 2025 P2
If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^2 A + \sin ^2 B + \sin^2  C = 2 \left( \cos ^2 A + \cos ^2 B + \cos ^2 C \right),$$prove that the triangle must have a right angle.
6 replies
SomeonecoolLovesMaths
Yesterday at 11:16 AM
quasar_lord
an hour ago
ISI UGB 2025 P6
SomeonecoolLovesMaths   3
N an hour ago by Shiny_zubat
Source: ISI UGB 2025 P6
Let $\mathbb{N}$ denote the set of natural numbers, and let $\left( a_i, b_i \right)$, $1 \leq i \leq 9$, be nine distinct tuples in $\mathbb{N} \times \mathbb{N}$. Show that there are three distinct elements in the set $\{ 2^{a_i} 3^{b_i} \colon 1 \leq i \leq 9 \}$ whose product is a perfect cube.
3 replies
SomeonecoolLovesMaths
Yesterday at 11:18 AM
Shiny_zubat
an hour ago
Shortest number theory you might've seen in your life
AlperenINAN   5
N an hour ago by Royal_mhyasd
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)$ is a perfect square, then $pq + 1$ is also a perfect square.
5 replies
AlperenINAN
Yesterday at 7:51 PM
Royal_mhyasd
an hour ago
d+2 pts in R^d can partition
EthanWYX2009   0
3 hours ago
Source: Radon's Theorem
Show that: any set of $d + 2$ points in $\mathbb R^d$ can be partitioned into two sets whose convex hulls intersect.
0 replies
EthanWYX2009
3 hours ago
0 replies
hard inequality omg
tokitaohma   4
N 3 hours ago by arqady
1. Given $a, b, c > 0$ and $abc=1$
Prove that: $ \sqrt{a^2+1} + \sqrt{b^2+1} + \sqrt{c^2+1} \leq \sqrt{2}(a+b+c) $

2. Given $a, b, c > 0$ and $a+b+c=1 $
Prove that: $ \dfrac{\sqrt{a^2+2ab}}{\sqrt{b^2+2c^2}} + \dfrac{\sqrt{b^2+2bc}}{\sqrt{c^2+2a^2}} + \dfrac{\sqrt{c^2+2ca}}{\sqrt{a^2+2b^2}} \geq \dfrac{1}{a^2+b^2+c^2} $
4 replies
tokitaohma
Yesterday at 5:24 PM
arqady
3 hours ago
Advanced topics in Inequalities
va2010   23
N Apr 22, 2025 by Novmath
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
23 replies
va2010
Mar 7, 2015
Novmath
Apr 22, 2025
Advanced topics in Inequalities
G H J
G H BBookmark kLocked kLocked NReply
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va2010
1276 posts
#1 • 11 Y
Y by DrMath, jh235, TheCrafter, Eugenis, mathleticguyyy, somebodyyouusedtoknow, Adventure10, Mango247, NicoN9, cubres, ehuseyinyigit
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
Attachments:
advanced-topics-inequalities (8).pdf (139kb)
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tastymath75025
3223 posts
#2 • 1 Y
Y by Adventure10
Will edit later maybe

4.1: sol

5.1: sol

[/hide]
This post has been edited 2 times. Last edited by tastymath75025, Mar 7, 2015, 5:31 AM
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Nuran2010
95 posts
#3 • 2 Y
Y by TunarHasanzade, TDVOLIMPTEAM
3.1:Click to reveal hidden text
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Nuran2010
95 posts
#4 • 3 Y
Y by TunarHasanzade, Strangett, TDVOLIMPTEAM
Similar logic,but a bit different approach for 5.1:
Click to reveal hidden text
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sqing
42122 posts
#5 • 1 Y
Y by Strangett
Problem 5.1. (Tuan Le)
Let $a,b,c$ be positive real numbers such that $  abc\geq 1$ Prove the inequality$$\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\le 1.$$
Problem4.2. (T.Q.Anh)
Let $a,b,c>0$ and $ab+bc+ca=3$.Prove that$$ \frac{a}{2a+b^2}+\frac{b}{2b+c^2}+\frac{c}{2c+a^2} \leq 1$$
Problem 3.1. (Titu Andresscu)
Let $ a,b,c>0 $ and $ a+b+c\geq 3 .$ Prove that$$ \frac{1}{a^2+b+c}+\frac{1}{b^2+c+a}+\frac{1}{c^2+a+b}\leq 1$$https://artofproblemsolving.com/community/c6h1838005p12333824
This post has been edited 2 times. Last edited by sqing, Apr 16, 2025, 2:33 PM
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giangtruong13
146 posts
#6
Y by
You see NOTHING
This post has been edited 1 time. Last edited by giangtruong13, Apr 16, 2025, 1:55 PM
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Primeniyazidayi
101 posts
#7
Y by
The direction of 3.2 is false.Anyway,here is an approach from "Problems from the book,Titu Andreescu and Gabriel Dospinescu"
S3.2
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sqing
42122 posts
#8
Y by
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$
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sqing
42122 posts
#9
Y by
Let $x,y,z>0$ and $xyz=x+y+z$.Prove that$$\frac{(x-1)^2}{x^2+1}+\frac{(y-1)^2}{y^2+1}+\frac{(z-1)^2}{z^2+1} \geq 3-\frac{3\sqrt 3}{2}$$*
Let $ a, b, c\geq -\frac{3}{4},a+b+c=1 .$ Prove that
$$\frac{(a-1)^{2}}{a^2+1 }+\frac{(b-1)^{2}}{b^2+1 }+\frac{(c-1)^{2}}{c^2+1 }\geq \frac{6}{5}$$Let $ a, b, c>0,abc\geq 8 .$ Prove that $$\dfrac{(a-1)^2}{a^2+2}+\dfrac{(b-1)^2}{b^2+2}+\dfrac{(c-1)^2}{c^2+2}\geq\frac{25}{38}$$
This post has been edited 1 time. Last edited by sqing, Apr 16, 2025, 3:27 PM
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Strangett
2 posts
#10 • 1 Y
Y by Nuran2010
Problem 3.4. Given that \( a + b + c = 3 \), prove that
\[
(2a^2 + 3)(2b^2 + 3)(2c^2 + 3) \ge 125.
\]
Click to reveal hidden text
This post has been edited 1 time. Last edited by Strangett, Apr 16, 2025, 5:18 PM
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Novmath
20 posts
#12 • 1 Y
Y by Strangett
Strangett wrote:
Problem 3.4. Given that \( a + b + c = 3 \), prove that
\[
(2a^2 + 3)(2b^2 + 3)(2c^2 + 3) \ge 125.
\]
Click to reveal hidden text

Nice solution ,can you explain how did you think of taking f(x)= something with ln. When do we take f(x) as ln(something) ?
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Novmath
20 posts
#13
Y by
va2010 wrote:
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!

The problem 2.1 (USAMO 2001 P3) is wrong. The condition was $a^2+b^2+c^2+ abc =4$
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Strangett
2 posts
#14
Y by
Novmath wrote:
Strangett wrote:
Problem 3.4. Given that \( a + b + c = 3 \), prove that
\[
(2a^2 + 3)(2b^2 + 3)(2c^2 + 3) \ge 125.
\]
Click to reveal hidden text

Nice solution ,can you explain how did you think of taking f(x)= something with ln. When do we take f(x) as ln(something) ?

Just in order to use the sum property of the natural logarithm (to apply Jensen's inequality to sums of function values). Also when taking the derivatives it is pretty obvious we are going to get something with its denominator squared and it is surely a convex function.
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sqing
42122 posts
#15
Y by
Novmath wrote:
va2010 wrote:
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!

The problem 2.1 (USAMO 2001 P3) is wrong. The condition was $a^2+b^2+c^2+ abc =4$
Yeah:
Let $a, b, c \geq 0$ and satisfy $ a^2+b^2+c^2 +abc = 4 . $ Show that\[ 0 \le ab + bc + ca - abc \leq 2. \]
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sqing
42122 posts
#16
Y by
Let $a, b, c \geq 0$ and $ a^2+b^2+c^2 +2abc = 4 . $ Show that$$ 0 \le ab + bc + ca - abc \leq 2$$
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Novmath
20 posts
#17
Y by
Strangett wrote:
Just in order to use the sum property of the natural logarithm (to apply Jensen's inequality to sums of function values). Also when taking the derivatives it is pretty obvious we are going to get something with its denominator squared and it is surely a convex function.
Thank you for explanation.
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Novmath
20 posts
#18
Y by
sqing wrote:
Novmath wrote:
va2010 wrote:
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!

The problem 2.1 (USAMO 2001 P3) is wrong. The condition was $a^2+b^2+c^2+ abc =4$
Yeah:
Let $a, b, c \geq 0$ and satisfy $ a^2+b^2+c^2 +abc = 4 . $ Show that\[ 0 \le ab + bc + ca - abc \leq 2. \]

Among a, b, c there must exist 2 numbers, say a and b, such that $ (1-a)(1-b) \geq 0$
Then: $ 1 \geq a + b - ab$
On the other hand, from $ a^2 + b^2 + c^2 + abc = 4$ we have:
$ c = \frac {-ab + \sqrt{a^2b^2 - 4a^2 - 4b^2 + 16}}{2} \leq \frac {-ab + \sqrt{a^2b^2 - 8ab + 16}}{2} = -ab +2$
Thus:
$ ab + bc + ca - abc = ab + c(a+b-ab) \leq ab + c \leq 2  (q.e.d)$
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Novmath
20 posts
#19
Y by
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
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sqing
42122 posts
#20
Y by
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474
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Novmath
20 posts
#21
Y by
sqing wrote:
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474

Thank you very much :coolspeak:
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Novmath
20 posts
#22
Y by
sqing wrote:
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474

Bro where can I learn separation trick in inequalities ,do you have some source which explaines it detailed, if you send it to me it will be appreciated!
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Novmath
20 posts
#23
Y by
sqing wrote:
Let $a, b, c \geq 0$ and $ a^2+b^2+c^2 +2abc = 4 . $ Show that$$ 0 \le ab + bc + ca - abc \leq 2$$

Can you also share solution of this problem:
(Vasc) Prove that if \( a, b, c \geq 0 \) and
\[
x = a + \frac{1}{b}, \quad y = b + \frac{1}{c}, \quad z = c + \frac{1}{a},
\]then
\[
xy + yz + zx \geq 2 + x + y + z.
\]
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Primeniyazidayi
101 posts
#24
Y by
Solution 3.3
This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 21, 2025, 8:34 PM
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Novmath
20 posts
#25
Y by
sqing wrote:
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474

Please share the solution of Vasc one that I shared
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