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Inequality
Amin12   8
N an hour ago by A.H.H
Source:  Iran 3rd round-2017-Algebra final exam-P3
Let $a,b$ and $c$ be positive real numbers. Prove that
$$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$
8 replies
Amin12
Sep 2, 2017
A.H.H
an hour ago
Not so beautiful
m4thbl3nd3r   0
an hour ago
Let $a, b,c>0$ such that $b+c>a$. Prove that $$2 \sqrt[4]{\frac{a}{b+c-a}}\ge 2 +\frac{2a^2-b^2-c^2}{(a+b)(a+c)}.$$
0 replies
m4thbl3nd3r
an hour ago
0 replies
Every subset of size k has sum at most N/2
orl   50
N 2 hours ago by de-Kirschbaum
Source: USAMO 2006, Problem 2, proposed by Dick Gibbs
For a given positive integer $k$ find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k + 1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $\tfrac{N}{2}.$
50 replies
orl
Apr 20, 2006
de-Kirschbaum
2 hours ago
Inspired by a9opsow_
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b > 0  .$ Prove that
$$ \frac{(ka^2 - kab-b)^2 + (kb^2 - kab-a)^2 + (ab-ka-kb )^2}{ (ka+b)^2 + (kb+a)^2+(a - b)^2 }\geq  \frac {1}{(k+1)^2}$$Where $ k\geq 0.37088 .$
$$\frac{(a^2 - ab-b)^2 + (b^2 - ab-a)^2 + ( ab-a-b)^2}{a^2 +b^2+(a - b)^2 } \geq 1$$$$ \frac{(2a^2 - 2ab-b)^2 + (2b^2 - 2ab-a)^2 + (ab-2a-2b )^2}{ (2a+b)^2 + (2b+a)^2+(a - b)^2 }\geq  \frac 19$$
2 replies
sqing
3 hours ago
sqing
2 hours ago
Cute NT Problem
M11100111001Y1R   5
N 3 hours ago by compoly2010
Source: Iran TST 2025 Test 4 Problem 1
A number \( n \) is called lucky if it has at least two distinct prime divisors and can be written in the form:
\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]where \( p_1, \dots, p_k \) are distinct prime numbers that divide \( n \). (Note: it is possible that \( n \) has other prime divisors not among \( p_1, \dots, p_k \).) Prove that for every prime number \( p \), there exists a lucky number \( n \) such that \( p \mid n \).
5 replies
M11100111001Y1R
Tuesday at 7:20 AM
compoly2010
3 hours ago
3 var inequality
SunnyEvan   13
N 4 hours 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
SunnyEvan
May 17, 2025
Nguyenhuyen_AG
4 hours ago
trigonometric inequality
MATH1945   13
N 4 hours 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
4 hours ago
Iran TST Starter
M11100111001Y1R   2
N 5 hours 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
5 hours ago
Twin Prime Diophantine
awesomeming327.   23
N 5 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
awesomeming327.
Mar 7, 2025
HDavisWashu
5 hours ago
Troublesome median in a difficult inequality
JG666   2
N 5 hours ago by navid
Source: 2022 Spring NSMO Day 2 Problem 3
Determine the minimum value of $\lambda\in\mathbb{R}$, such that for any positive integer $n$ and non-negative reals $x_1, x_2, \cdots, x_n$, the following inequality always holds:
$$\sum_{i=1}^n(m_i-a_i)^2\leqslant \lambda\cdot\sum_{i=1}^nx_i^2,$$Here $m_i$ and $a_i$ denote the median and arithmetic mean of $x_1, x_2, \cdots, x_i$, respectively.

Duanyang ZHANG, High School Affiliated to Renmin University of China
2 replies
JG666
May 22, 2022
navid
5 hours ago
beautiful functional equation problem
Medjl   6
N Apr 6, 2025 by Sadigly
Source: Netherlands TST for BxMO 2017 problem 2
Let define a function $f: \mathbb{N} \rightarrow \mathbb{Z}$ such that :
$i)$$f(p)=1$ for all prime numbers $p$.
$ii)$$f(xy)=xf(y)+yf(x)$ for all positive integers $x,y$
find the smallest $n \geq 2016$ such that $f(n)=n$
6 replies
Medjl
Feb 1, 2018
Sadigly
Apr 6, 2025
beautiful functional equation problem
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G H BBookmark kLocked kLocked NReply
Source: Netherlands TST for BxMO 2017 problem 2
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Medjl
757 posts
#1 • 4 Y
Y by Muradjl, Adventure10, Mango247, PikaPika999
Let define a function $f: \mathbb{N} \rightarrow \mathbb{Z}$ such that :
$i)$$f(p)=1$ for all prime numbers $p$.
$ii)$$f(xy)=xf(y)+yf(x)$ for all positive integers $x,y$
find the smallest $n \geq 2016$ such that $f(n)=n$
This post has been edited 2 times. Last edited by Medjl, Feb 1, 2018, 3:16 PM
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Math-Ninja
3749 posts
#2 • 2 Y
Y by Adventure10, PikaPika999
Why is this named beautiful problem. Titles are meant for description please.
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ThE-dArK-lOrD
4071 posts
#3 • 7 Y
Y by rkm0959, Muradjl, Math-Ninja, Com10atorics, Adventure10, Mango247, PikaPika999
Not hard to prove by induction on $s\in \mathbb{Z}^+$ that $$f(n)=n\sum_{i=1}^{k}{\frac{\alpha_i }{p_i}}$$where $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ is the canonical form of $n$ with $\sum_{i=1}^{k}{\alpha_i} =s$.
Hence, $f(n)=n\sum_{i=1}^{k}{\frac{\alpha_i }{p_i}}$ where $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ is the canonical form of $n$ is true for all positive integer $n$.
We want to find $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ that $f(n)=n\iff \sum_{i=1}^{k}{\frac{\alpha_i }{p_i}} =1$.

If $k\geq 2$, we get that $\alpha_i \leq p_i$ for all $i\in \{ 1,2,...,k\}$.
But we've $\sum_{i=1}^{k}{\alpha_i \times \frac{P}{p_i} }=P$ where $P=\prod_{i=1}^{k}{p_i}$. Modulo $p_1$ gives us $p_1\mid \alpha_1 \times \frac{P}{p_1}$, impossible since all $p_i$s are distinct.

So, $k=1$. This gives $n=p^p$ for some prime number $p$.
Easy to see that the smallest such $n$ that not smaller than $2016$ is $3125=5^5$.
This post has been edited 1 time. Last edited by ThE-dArK-lOrD, Feb 1, 2018, 4:58 PM
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soryn
5348 posts
#4 • 2 Y
Y by Adventure10, PikaPika999
Very nice! I like this problem, I like this solution! Congratulations!:))))))
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User335559
472 posts
#5 • 3 Y
Y by Adventure10, Mango247, PikaPika999
ThE-dArK-lOrD wrote:
Not hard to prove by induction on $s\in \mathbb{Z}^+$ that $$f(n)=n\sum_{i=1}^{k}{\frac{\alpha_i }{p_i}}$$where $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ is the canonical form of $n$ with $\sum_{i=1}^{k}{\alpha_i} =s$.
Hence, $f(n)=n\sum_{i=1}^{k}{\frac{\alpha_i }{p_i}}$ where $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ is the canonical form of $n$ is true for all positive integer $n$.
We want to find $n=\prod_{i=1}^{k}{p_i^{\alpha_i}}$ that $f(n)=n\iff \sum_{i=1}^{k}{\frac{\alpha_i }{p_i}} =1$.

If $k\geq 2$, we get that $\alpha_i \leq p_i$ for all $i\in \{ 1,2,...,k\}$.
But we've $\sum_{i=1}^{k}{\alpha_i \times \frac{P}{p_i} }=P$ where $P=\prod_{i=1}^{k}{p_i}$. Modulo $p_1$ gives us $p_1\mid \alpha_1 \times \frac{P}{p_1}$, impossible since all $p_i$s are distinct.

So, $k=1$. This gives $n=p^p$ for some prime number $p$.
Easy to see that the smallest such $n$ that not smaller than $2016$ is $3125=5^5$.

How did you find the function? What's the motivation behind that?
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liekkas
370 posts
#6 • 3 Y
Y by Adventure10, Mango247, PikaPika999
It's 2015 Austria MO (maybe not posted yet)
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Sadigly
229 posts
#7 • 1 Y
Y by PikaPika999
Easy to see that $f(0)=0$ and $f(1)=0$, and them being the only zeroes of this function. One can also say $f$ doesn't take any negative numbers.

Induct to get $f(p^x)=xp^{x-1}$

Let $n=p^x\cdot A$, where $p\nmid A$, and assume this number satisfies the given condition

$p^x\cdot A=n=f(n)=f(p^x\cdot A)=p^xf(A)+Af(p^x)=p^xf(A)+xAp^{x-1}$

$LHS\equiv 0~(mod~p^x)\Rightarrow RHS\equiv0~(mod~p^x).$ We get that $p\mid x$. Rewriting $x=p\cdot k$ gives $$(1-k)A=f(A)$$It should be obvious that $k=1$ and $f(A)=0$. Since $A\neq0$ (Or else $n=0$), we have $n=p^p$. All that is left to find the smallest prime that satisfies $p^p\geq 2016$, which is $5$
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