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Help me this problem. Thank you
illybest   3
N an hour ago by jasperE3
Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z
3 replies
illybest
Today at 11:05 AM
jasperE3
an hour ago
line JK of intersection points of 2 lines passes through the midpoint of BC
parmenides51   4
N 2 hours ago by reni_wee
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 $BC$
4 replies
parmenides51
Dec 11, 2018
reni_wee
2 hours ago
AGM Problem(Turkey JBMO TST 2025)
HeshTarg   3
N 2 hours ago by ehuseyinyigit
Source: Turkey JBMO TST Problem 6
Given that $x, y, z > 1$ are real numbers, find the smallest possible value of the expression:
$\frac{x^3 + 1}{(y-1)(z+1)} + \frac{y^3 + 1}{(z-1)(x+1)} + \frac{z^3 + 1}{(x-1)(y+1)}$
3 replies
HeshTarg
3 hours ago
ehuseyinyigit
2 hours ago
Shortest number theory you might've seen in your life
AlperenINAN   8
N 2 hours ago by HeshTarg
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
8 replies
AlperenINAN
Yesterday at 7:51 PM
HeshTarg
2 hours ago
Squeezing Between Perfect Squares and Modular Arithmetic(JBMO TST Turkey 2025)
HeshTarg   3
N 2 hours ago by BrilliantScorpion85
Source: Turkey JBMO TST Problem 4
If $p$ and $q$ are primes and $pq(p+1)(q+1)+1$ is a perfect square, prove that $pq+1$ is a perfect square.
3 replies
HeshTarg
3 hours ago
BrilliantScorpion85
2 hours ago
A bit too easy for P2(Turkey 2025 JBMO TST)
HeshTarg   0
2 hours ago
Source: Turkey 2025 JBMO TST P2
Let $n$ be a positive integer. Aslı and Zehra are playing a game on an $n \times n$ chessboard. Initially, $10n^2$ stones are placed on the squares of the board. In each move, Aslı chooses a row or a column; Zehra chooses a row or a column. The number of stones in each square of the chosen row or column must change such that the difference between the number of stones in a square with the most stones and a square with the fewest stones in that same row or column is at most 1. For which values of $n$ can Aslı guarantee that after a finite number of moves, all squares on the board will have an equal number of stones, regardless of the initial distribution?
0 replies
HeshTarg
2 hours ago
0 replies
D1030 : An inequalitie
Dattier   0
2 hours ago
Source: les dattes à Dattier
Let $0<a<b<c<d$ reals, and $n \in \mathbb N^*$.

Is it true that $a^n(b-a)+b^n(c-b)+c^n(d-c) \leq \dfrac {d^{n+1}}{n+1}$ ?
0 replies
Dattier
2 hours ago
0 replies
Long and wacky inequality
Royal_mhyasd   0
3 hours ago
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
0 replies
Royal_mhyasd
3 hours ago
0 replies
Vietnamese national Olympiad 2007, problem 4
hien   16
N 3 hours ago by de-Kirschbaum
Given a regular 2007-gon. Find the minimal number $k$ such that: Among every $k$ vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.
16 replies
hien
Feb 8, 2007
de-Kirschbaum
3 hours ago
2n^2+4n-1 and 3n+4 have common powers
bin_sherlo   4
N 3 hours ago by CM1910
Source: Türkiye 2025 JBMO TST P5
Find all positive integers $n$ such that a positive integer power of $2n^2+4n-1$ equals to a positive integer power of $3n+4$.
4 replies
bin_sherlo
Yesterday at 7:13 PM
CM1910
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
42130 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
104 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
42130 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
42130 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
42130 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
42130 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
42130 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
104 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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