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inequalities
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Very easy inequality
pggp   2
N 12 minutes ago by ali123456
Source: Polish Junior MO Second Round 2019
Let $x$, $y$ be real numbers, such that $x^2 + x \leq y$. Prove that $y^2 + y \geq x$.
2 replies
pggp
Oct 26, 2020
ali123456
12 minutes ago
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Problem about Euler's function
luutrongphuc   1
N 13 minutes ago by ishan.panpaliya
Prove that for every integer $n \ge 5$, we have:
$$ 2^{n^2+3n-13} \mid \phi \left(2^{2^{n}}-1 \right)$$
1 reply
luutrongphuc
2 hours ago
ishan.panpaliya
13 minutes ago
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Problem 5
blug   1
N 34 minutes ago by WallyWalrus
Source: Polish Junior Math Olympiad Finals 2025
Each square on a 5×5 board contains an arrow pointing up, down, left, or right. Show that it is possible to remove exactly 20 arrows from this board so that no two of the remaining five arrows point to the same square.
1 reply
blug
Mar 15, 2025
WallyWalrus
34 minutes ago
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Cool Number Theory
Fermat_Fanatic108   6
N an hour ago by epl1
For an integer with 5 digits $n=abcde$ (where $a, b, c, d, e$ are the digits and $a\neq 0$) we define the \textit{permutation sum} as the value $$bcdea+cdeab+deabc+eabcd$$For example the permutation sum of 20253 is $$02532+25320+53202+32025=113079$$Let $m$ and $n$ be two fivedigit integers with the same permutation sum.
Prove that $m=n$.
6 replies
Fermat_Fanatic108
5 hours ago
epl1
an hour ago
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Incenter geometry with parallel lines
nAalniaOMliO   1
N an hour ago by LenaEnjoyer
Source: Belarusian MO 2023
Let $\omega$ be the incircle of triangle $ABC$. Line $l_b$ is parallel to side $AC$ and tangent to $\omega$. Line $l_c$ is parallel to side $BC$ and tangent to $\omega$. It turned out that the intersection point of $l_b$ and $l_c$ lies on circumcircle of $ABC$
Find all possible values of $\frac{AB+AC}{BC}$
1 reply
nAalniaOMliO
Apr 16, 2024
LenaEnjoyer
an hour ago
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Problem for VASC, SI Book
hungkhtn   21
N 2 hours ago by imnotgoodatmathsorry
Source: please let him prove it first
Let $a,b,c$ be non-negative real numbers such that $a+b+c=3$. Prove that
\[a\sqrt{1+b^{3}}+b\sqrt{1+c^{3}}+c\sqrt{1+a^{3}}\le 5.\]
21 replies
hungkhtn
Jun 5, 2007
imnotgoodatmathsorry
2 hours ago
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IMO Shortlist 2009 - Problem N4
April   12
N 2 hours ago by asdf334
Find all positive integers $n$ such that there exists a sequence of positive integers $a_1$, $a_2$,$\ldots$, $a_n$ satisfying: \[a_{k+1}=\frac{a_k^2+1}{a_{k-1}+1}-1\] for every $k$ with $2\leq k\leq n-1$.

Proposed by North Korea
12 replies
April
Jul 5, 2010
asdf334
2 hours ago
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China Team Selection Test 2015 TST 1 Day 2 Q1
sqing   6
N 2 hours ago by sttsmet
Source: China Hangzhou
Prove that : For each integer $n \ge 3$, there exists the positive integers $a_1<a_2< \cdots <a_n$ , such that for $ i=1,2,\cdots,n-2 $ , With $a_{i},a_{i+1},a_{i+2}$ may be formed as a triangle side length , and the area of the triangle is a positive integer.
6 replies
sqing
Mar 14, 2015
sttsmet
2 hours ago
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China Mathematical Olympiad 1993 problem5
jred   3
N 2 hours ago by iStud
Source: China Mathematical Olympiad 1993 problem5
$10$ students bought some books in a bookstore. It is known that every student bought exactly three kinds of books, and any two of them shared at least one kind of book. Determine, with proof, how many students bought the most popular book at least? (Note: the most popular book means most students bought this kind of book)
3 replies
jred
Sep 23, 2013
iStud
2 hours ago
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x and o game, in an infinite grid of regular triangles
parmenides51   5
N 3 hours ago by Lil_flip38
Source: Norwegian Mathematical Olympiad 2017 - Abel Competition p3b
In an infinite grid of regular triangles, Niels and Henrik are playing a game they made up.
Every other time, Niels picks a triangle and writes $\times$ in it, and every other time, Henrik picks a triangle where he writes a $o$. If one of the players gets four in a row in some direction (see figure), he wins the game.
Determine whether one of the players can force a victory.
IMAGE
5 replies
parmenides51
Sep 3, 2019
Lil_flip38
3 hours ago
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BMN is equilateral iff rectangle ABCD is square
parmenides51   4
N 3 hours ago by Tsikaloudakis
Source: 2004 Romania NMO SL - Shortlist VII-VIII p8 https://artofproblemsolving.com/community/c3950157_
Consider a point $M$ on the diagonal $BD$ of a given rectangle $ABCD$, such that $\angle AMC = \angle CMD$. The point $N$ is the intersection point between $AM$ and the parallel line to $CM$ that contains $B$. Prove that the triangle $BMN$ is equilateral if and only if $ABCD$ is a square.

Valentin Vornicu
4 replies
parmenides51
Sep 16, 2024
Tsikaloudakis
3 hours ago
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an inequality
BuiBaAnh   11
N Jun 2, 2020 by BestChoice123
Let $a,b,c \geq 0$. Prove that:
$\frac{a(b+c)}{a^2+2bc}+\frac{b(c+a)}{b^2+2ac}+\frac{c(a+b)}{c^2+2ab} \geq 1+\frac{ab+ac+bc}{a^2+b^2+c^2}$



Hoa nhài
11 replies
BuiBaAnh
May 4, 2015
BestChoice123
Jun 2, 2020
J
an inequality
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inequalities
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BuiBaAnh
143 posts
BuiBaAnh
#1 h  May 4, 2015, 1:29 PM • 2 Y
Y by Adventure10, Mango247
Let $a,b,c \geq 0$. Prove that:
$\frac{a(b+c)}{a^2+2bc}+\frac{b(c+a)}{b^2+2ac}+\frac{c(a+b)}{c^2+2ab} \geq 1+\frac{ab+ac+bc}{a^2+b^2+c^2}$



Hoa nhài
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Wangzu
283 posts
Wangzu
#2 h  May 5, 2015, 8:16 AM • 3 Y
Y by A_Gappus, Adventure10, Mango247
We have the ineq for all non-negative:

$\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\geq \frac{2}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}$
$\Leftrightarrow (ab+bc+ac)(\sum \frac{1}{a^2+2bc})\geq 2+\frac{ab+bc+ac}{a^2+b^2+c^2}$
$\Leftrightarrow \sum \frac{a(b+c)}{a^2+2bc}+\frac{bc}{a^2+2bc}\geq 2+\frac{ab+bc+ac}{a^2+b^2+c^2} $

And then we have to prove the ineq:
$\sum \frac{bc}{a^2+2bc}\leq 1\Leftrightarrow \sum \frac{a^2}{a^2+2bc}\geq 1$ which just by C-S
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Wangzu
283 posts
Wangzu
#3 h  May 5, 2015, 8:24 AM • 2 Y
Y by Adventure10, Mango247
I have another idea by S.O.S, our inequality equivalent to:

$S_{a}(b-c)^2+S_{b}(a-c)^2+S_{c}(a-b)^2\geq 0$

And $S_{a}=\frac{3}{2(a^2+b^2+c^2)}-\frac{4ab+4ac-4a^2-bc}{(b^2+2ac)(c^2+2ab)}$
$S_{b}=\frac{3}{2(a^2+b^2+c^2)}-\frac{4bc+4ab-4b^2-ac}{(a^2+2bc)(c^2+2ab)}$
$S_{c}=\frac{3}{2(a^2+b^2+c^2)}-\frac{4ac+4bc-4c^2-ab}{(a^2+2bc)(b^2+2ac)}$

Can someone prove this(by this idea)?
This post has been edited 1 time. Last edited by Wangzu, May 5, 2015, 8:29 AM
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xzlbq
15849 posts
xzlbq
#4 h  May 5, 2015, 9:56 AM • 4 Y
Y by Wangzu, dragonheart6, Adventure10, Mango247
BuiBaAnh wrote:
Let $a,b,c \geq 0$. Prove that:
$F=\frac{a(b+c)}{a^2+2bc}+\frac{b(c+a)}{b^2+2ac}+\frac{c(a+b)}{c^2+2ab}-1-\frac{ab+ac+bc}{a^2+b^2+c^2}\geq 0$



Hoa nhài

a SOS is:

$f={\frac {Q}{ \left( {c}^{2}+2\,ab \right) \left( {b}^{2}+2\,ac \right) \left( {a}^{2}+2\,bc \right) \left( {a}^{2}+{b}^{2}+{c}^{2} \right) }}$

$Q= \left( 2\,a-b-c \right) ^{2}a{b}^{2}{c}^{3}+ \left( 2\,b-a-c \right) ^{2}b{c}^{2}{a}^{3}+ \left( 2\,c-a-b \right) ^{2}c{a}^{2}{b}^ {3}+1/2\, \left( -a{b}^{2}+{c}^{2}a-abc+{b}^{2}c \right) ^{2}{a}^{2}+1 /2\, \left( {a}^{2}b-b{c}^{2}-abc+{c}^{2}a \right) ^{2}{b}^{2}+1/2\, \left( {a}^{2}b-abc-{a}^{2}c+{b}^{2}c \right) ^{2}{c}^{2}+ \left( c-b \right) ^{2} \left( a-b \right) ^{2}{c}^{2}ab+ \left( b-a \right) ^{2 } \left( c-a \right) ^{2}{b}^{2}ca+ \left( a-c \right) ^{2} \left( b-c \right) ^{2}{a}^{2}bc+1/2\, \left( a-b \right) ^{2} \left( a+b-2\,c \right) ^{2}{c}^{4}+1/2\, \left( b-c \right) ^{2} \left( b-2\,a+c \right) ^{2}{a}^{4}+1/2\, \left( c-a \right) ^{2} \left( a-2\,b+c \right) ^{2}{b}^{4}$

BQ
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Wangzu
283 posts
Wangzu
#5 h  May 5, 2015, 10:36 AM • 2 Y
Y by Adventure10, Mango247
I have found this following inequality(may be easy):

For all positive real numbers,we have:

$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 1+\frac{16abc}{(a+b)(b+c)(a+c)}$
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szl6208
2032 posts
szl6208
#6 h  May 5, 2015, 11:48 AM • 2 Y
Y by Adventure10, Mango247
For #5
We have
$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}- [1+\frac{16abc}{(a+b)(b+c)(a+c)}]=$
$\sum{\frac{(a-c)^2(b-c)^2}{(c^2+ab)(b+c)(c+a)}}\ge{0}$
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lumia
93 posts
lumia
#7 h  May 5, 2015, 2:59 PM • 1 Y
Y by Adventure10
Wangzu wrote:
I have found this following inequality(may be easy):

For all positive real numbers,we have:

$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 1+\frac{16abc}{(a+b)(b+c)(a+c)}$
hello, I know it has a solution by using Am-Gm in a book of VQBCan and TQAnh :)
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arqady
30150 posts
arqady
#8 h  May 23, 2015, 9:40 AM • 3 Y
Y by luofangxiang, Adventure10, Mango247
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 2+\frac{3abc(a+b+c)}{(a^2+b^2+c^2)^2}$$
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arqady
30150 posts
arqady
#9 h  May 23, 2015, 10:03 AM • 2 Y
Y by Adventure10, Mango247
Wangzu wrote:
For all positive real numbers,we have:
$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 1+\frac{16abc}{(a+b)(b+c)(a+c)}$
$\sum_{cyc}\frac{a(b+c)}{a^2+bc}-1-\frac{16abc}{(a+b)(b+c)(a+c)}=2-\frac{16abc}{(a+b)(b+c)(a+c)}-\sum_{cyc}\left(1-\frac{a(b+c)}{a^2+bc}\right)=$
$=\frac{2\sum\limits_{cyc}(a^2b+a^2c-2abc)}{(a+b)(a+c)(b+c)}-\sum_{cyc}\frac{(a-b)(a-c)}{a^2+bc}=$
$=\frac{2\sum\limits_{cyc}(b+c)(a-b)(a-c)}{(a+b)(a+c)(b+c)}-\sum_{cyc}\frac{(a-b)(a-c)}{a^2+bc}=\sum_{cyc}\frac{(a-b)^2(a-c)^2}{(a+b)(a+c)(a^2+bc)}\geq0$.
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szl6208
2032 posts
szl6208
#10 h  May 24, 2015, 2:32 AM • 1 Y
Y by Adventure10
For #8
We have
\[[\sum{\frac{a(b+c)}{a^2+bc}}-2][\sum{a^2}]^2-3abc\sum{a}=\sum{\frac{ab(2a^2+ab+2b^2)(ab+c^2)(a-b)^2}{2(a^2+bc)(ac+b^2)}}\]
\[+\frac{(a-b)^2(b-c)^2(c-a)^2}{2(a^2+bc)(ac+b^2)(ab+c^2)}\sum{(a^2-ab+b^2)^2}\ge{0}\]
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luofangxiang
4613 posts
luofangxiang
#11 h  Nov 30, 2016, 10:08 AM • 2 Y
Y by teomihai, Adventure10
arqady wrote:
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 2+\frac{3abc(a+b+c)}{(a^2+b^2+c^2)^2}$$

//cdn.artofproblemsolving.com/images/9/7/0/97089c0e44943dbe33c6f1d7a85c6aceaaa41e66.jpg
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BestChoice123
1119 posts
BestChoice123
#12 h  Jun 2, 2020, 2:19 PM • 2 Y
Y by teomihai, Mango247
arqady wrote:
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 2+\frac{3abc(a+b+c)}{(a^2+b^2+c^2)^2}$$

$$LHS-RHS=\frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2+bc)(b^2+ca)(c^2+ab)}+\frac{abc[\sum ab(a+b)-abc]\sum (a^2-b^2)^2}{2(a^2+bc)(b^2+ca)(c^2+ab)(a^2+b^2+c^2)^2}\ge 0$$
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