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minimum of \sqrt{\frac{a}{b(3a+2)}}+\sqrt{\frac{b}{a(3b+2)}}
parmenides51   11
N 28 minutes ago by sqing
Source: JBMO Shortlist 2017 A2
Let $a$ and $b$ be positive real numbers such that $3a^2 + 2b^2 = 3a + 2b$. Find the minimum value of $A =\sqrt{\frac{a}{b(3a+2)}} + \sqrt{\frac{b}{a(2b+3)}} $
11 replies
parmenides51
Jul 25, 2018
sqing
28 minutes ago
IMO Genre Predictions
ohiorizzler1434   14
N 31 minutes ago by BR1F1SZ
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
14 replies
ohiorizzler1434
Today at 6:51 AM
BR1F1SZ
31 minutes ago
Inequality
MathsII-enjoy   0
44 minutes ago
A interesting problem generalized :-D
0 replies
MathsII-enjoy
44 minutes ago
0 replies
Interesting inequalities
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b,c>0 $ and $ (a+b)^2 (a+c)^2=16abc. $ Prove that
$$ 2a+b+c\leq \frac{128}{27}$$$$ \frac{9}{2}a+b+c\leq \frac{864}{125}$$$$3a+b+c\leq 24\sqrt{3}-36$$$$5a+b+c\leq \frac{4(8\sqrt{6}-3)}{9}$$
1 reply
sqing
Yesterday at 2:35 PM
sqing
an hour ago
Geometry Problem
Euler_Gauss   0
an hour ago
Given that $D$ is the midpoint of $BC$, $DM$ bisects $\angle ADB$ and intersects $AB$ at $M$, $I$ is the incenter of $\triangle {}{}{}ABD$, $AT$ bisects $\angle BAC$ and intersects the circumcircle of \(\triangle {}{}ABC\) at $T$, $MS$ is parallel to $BC$ and intersects $AT$ at $S$. Prove that $\angle MIS + \angle BIT = \pi.$
0 replies
Euler_Gauss
an hour ago
0 replies
Tricky invariant with 3 numbers on the board
Nuran2010   0
an hour ago
Source: Azerbaijan Junior National Olympiad 2021
Initially, the numbers $1,1,-1$ written on the board.At every step,Mikail chooses the two numbers $a,b$ and substitutes them with $2a+c$ and $\frac{b-c}{2}$ where $c$ is the unchosen number on the board. Prove that at least $1$ negative number must be remained on the board at any step.
0 replies
Nuran2010
an hour ago
0 replies
Inequality with 7th degree root.
Nuran2010   1
N an hour ago by sqing
Source: Azerbaijan Junior National Olympiad 2021 P3
For $a,b,c>0$ positive real numbers,prove the inequality:

$\sqrt[7]{\frac{a}{b+c}+\frac{b}{a+c}}+\sqrt[7]{\frac{b}{a+c}+\frac{c}{a+b}}+\sqrt[7]{\frac{c}{a+b}+\frac{a}{b+c}} \geq 3$.
1 reply
Nuran2010
an hour ago
sqing
an hour ago
A coincidence about triangles with common incenter
flower417477   5
N an hour ago by mashumaro
$\triangle ABC,\triangle ADE$ have the same incenter $I$.Prove that $BCDE$ is concyclic iff $BC,DE,AI$ is concurrent
5 replies
flower417477
Apr 30, 2025
mashumaro
an hour ago
Showing a certain number is divisible by 13
BBNoDollar   0
an hour ago
Source: Romanian Mathematical Gazette 2025
Show that 3^(n+2) + 9^(n+1) + 4^(2n+1) + 4^(4n+1) is divisible by 13 for every n natural number.
0 replies
BBNoDollar
an hour ago
0 replies
|a^2-b^2-2abc|<2c implies abc EVEN!
tom-nowy   0
an hour ago
Source: Own
Prove that if integers $a, b$ and $c$ satisfy $\left| a^2-b^2-2abc \right| <2c $, then $abc$ is an even number.
0 replies
tom-nowy
an hour ago
0 replies
inequalities
Tamako22   1
N an hour ago by sqing
let $a,b,c> 1,\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=1.$
prove that$$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \dfrac{2}{\sqrt{a}}+\dfrac{2}{\sqrt{b}}+\dfrac{2}{\sqrt{c}}$$
1 reply
Tamako22
2 hours ago
sqing
an hour ago
interesting function equation (fe) in IR
skellyrah   2
N Apr 23, 2025 by jasperE3
Source: mine
find all function F: IR->IR such that $$ xf(f(y)) + yf(f(x)) = f(xf(y)) + f(xy) $$
2 replies
skellyrah
Apr 23, 2025
jasperE3
Apr 23, 2025
interesting function equation (fe) in IR
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Source: mine
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skellyrah
23 posts
#1
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find all function F: IR->IR such that $$ xf(f(y)) + yf(f(x)) = f(xf(y)) + f(xy) $$
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CrazyInMath
457 posts
#2
Y by
solution
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jasperE3
11288 posts
#4
Y by
skellyrah wrote:
find all function F: IR->IR such that $$ xf(f(y)) + yf(f(x)) = f(xf(y)) + f(xy) $$

If IR means irrational numbers, and the problem is to find all $f$ such that $xf(f(y)) + yf(f(x)) = f(xf(y)) + f(xy)$ for all $x,y\in\mathbb{IR}$ then setting $x=f(y)^{-1}$ gives a contradiction
This post has been edited 1 time. Last edited by jasperE3, Apr 23, 2025, 9:43 PM
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