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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
inequality problem
pennypc123456789   2
N 4 minutes ago by pennypc123456789
Given $a,b,c$ be positive real numbers . Prove that
$$\frac{ab}{(a+b)^2} +\frac{bc}{(b+c)^2}+\frac{ac}{(a+c)^2} \ge \frac{6abc }{(a+b)(b+c)(a+c)}$$
2 replies
pennypc123456789
Yesterday at 2:42 PM
pennypc123456789
4 minutes ago
weird Condition
B1t   6
N 4 minutes ago by Assassino9931
Source: Mongolian TST 2025 P4
In triangle \(ABC\), where \(AC < AB\), the internal angle bisectors of angles \(\angle A\), \(\angle B\), and \(\angle C\) meet the sides \(BC\), \(AC\), and \(AB\) at points \(D\), \(E\), and \(F\), respectively. Let \( I \) be the incenter of triangle \( AEF \), and let \( G \) be the foot of the perpendicular from \( I \) to line \( BC \). Prove that if the quadrilateral \( DGEF \) is cyclic, then the center of its circumcircle lies on segment \( AD \).
6 replies
+1 w
B1t
Apr 27, 2025
Assassino9931
4 minutes ago
RMM 2019 Problem 2
math90   79
N 43 minutes ago by lpieleanu
Source: RMM 2019
Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is tangent to $\Omega$.

Jakob Jurij Snoj, Slovenia
79 replies
math90
Feb 23, 2019
lpieleanu
43 minutes ago
Good divisors and special numbers.
Nuran2010   2
N an hour ago by BR1F1SZ
Source: Azerbaijan Al-Khwarizmi IJMO TST 2024
$N$ is a positive integer. Call all positive divisors of $N$ which are different from $1$ and $N$ beautiful divisors.We call $N$ a special number when it has at least $2$ beautiful divisors and difference of any $2$ beautiful divisors divides $N$ as well. Find all special numbers.
2 replies
Nuran2010
Yesterday at 4:52 PM
BR1F1SZ
an hour ago
Inequality with 3 variables and a special condition
Nuran2010   2
N an hour ago by Assassino9931
Source: Azerbaijan Al-Khwarizmi IJMO TST 2024
For positive real numbers $a,b,c$ we have $3abc \geq ab+bc+ca$.
Prove that:

$\frac{1}{a^3+b^3+c}+\frac{1}{b^3+c^3+a}+\frac{1}{c^3+a^3+b} \leq \frac{3}{a+b+c}$.

Determine the equality case.
2 replies
Nuran2010
Yesterday at 5:06 PM
Assassino9931
an hour ago
Circumcircle of one triangle passes from another's circumcenter.
Nuran2010   1
N 2 hours ago by Assassino9931
Source: Azerbaijan Al-Khwarizmi IJMO TST 2024
In a parallelogram $ABCD$,$\angle A<90^\circ$ and $AB<BC$. Interior angle bisector of $\angle BAD$ intersects $BC$ at $M$, and $DC$ at $N$.Prove that circumcircle of $BCD$ passes from circumcenter of $CMN$.
1 reply
Nuran2010
Yesterday at 4:57 PM
Assassino9931
2 hours ago
a_i/i sequence
pad   19
N 2 hours ago by ihatemath123
Source: TSTST 2021/2
Let $a_1<a_2<a_3<a_4<\cdots$ be an infinite sequence of real numbers in the interval $(0,1)$. Show that there exists a number that occurs exactly once in the sequence
\[ \frac{a_1}{1},\frac{a_2}{2},\frac{a_3}{3},\frac{a_4}{4},\ldots.\]
Merlijn Staps
19 replies
pad
Nov 8, 2021
ihatemath123
2 hours ago
Find points with sames integer distances as given
nAalniaOMliO   2
N 2 hours ago by nAalniaOMliO
Source: Belarus TST 2024
Points $A_1, \ldots A_n$ with rational coordinates lie on a plane. It turned out that the distance between every pair of points is an integer. Prove that there exist points $B_1, \ldots ,B_n$ with integer coordinates such that $A_iA_j=B_iB_j$ for every pair $1 \leq i \leq j \leq n$
N. Sheshko, D. Zmiaikou
2 replies
nAalniaOMliO
Jul 17, 2024
nAalniaOMliO
2 hours ago
Geometry tangent circles
Stefan4024   68
N 2 hours ago by zuat.e
Source: EGMO 2016 Day 2 Problem 4
Two circles $\omega_1$ and $\omega_2$, of equal radius intersect at different points $X_1$ and $X_2$. Consider a circle $\omega$ externally tangent to $\omega_1$ at $T_1$ and internally tangent to $\omega_2$ at point $T_2$. Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega$.
68 replies
Stefan4024
Apr 13, 2016
zuat.e
2 hours ago
My Unsolved Problem
MinhDucDangCHL2000   2
N 3 hours ago by hukilau17
Source: 2024 HSGS Olympiad
Let triangle $ABC$ be inscribed in the circle $(O)$. A line through point $O$ intersects $AC$ and $AB$ at points $E$ and $F$, respectively. Let $P$ be the reflection of $E$ across the midpoint of $AC$, and $Q$ be the reflection of $F$ across the midpoint of $AB$. Prove that:
a) the reflection of the orthocenter $H$ of triangle $ABC$ across line $PQ$ lies on the circle $(O)$.
b) the orthocenters of triangles $AEF$ and $HPQ$ coincide.

Im looking for a solution used complex bashing :(
2 replies
MinhDucDangCHL2000
Yesterday at 4:53 PM
hukilau17
3 hours ago
Easy Combinatorics
MuradSafarli   2
N 3 hours ago by Sadigly
A student firstly wrote $x=3$ on the board. For each procces, the stutent deletes the number x and replaces it with either $(2x+4)$ or $(3x+8)$ or $(x^2+5x)$. Is this possible to make the number $(20^{25}+2024)$ on the board?
2 replies
MuradSafarli
5 hours ago
Sadigly
3 hours ago
4 variables with quadrilateral sides 2
mihaig   0
3 hours ago
Source: Own
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$\left(a+b+c+d-2\right)^2+8\geq3\left(abc+abd+acd+bcd\right).$$
0 replies
mihaig
3 hours ago
0 replies
Number theory
MuradSafarli   1
N 4 hours ago by Sadigly
Prove that for any natural number \( n \) :

\[
1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n + 1) \mid (4n + 3)(4n + 5) \cdot \ldots \cdot (8n + 3).
\]
1 reply
MuradSafarli
4 hours ago
Sadigly
4 hours ago
D1025 : Can you do that?
Dattier   0
4 hours ago
Source: les dattes à Dattier
Let $x_{n+1}=x_n^3$ and $x_0=3$.

Can you calculate $\sum\limits_{i=1}^{2^{2025}} x_i \mod 10^{30}$?
0 replies
Dattier
4 hours ago
0 replies
A simple inequality for highschool students
can_hang2007   37
N Nov 12, 2021 by sqing
Let $a ,\; b,\; c$ be positive real numbers such that
$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1.$
Prove that
$b^2+3 \ge 2b(a+c).$
37 replies
can_hang2007
Nov 3, 2010
sqing
Nov 12, 2021
A simple inequality for highschool students
G H J
G H BBookmark kLocked kLocked NReply
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can_hang2007
2948 posts
#1 • 7 Y
Y by xyzz, ZHEKSHEN, a_friendwr_a, son7, Ha_ha_ha, Adventure10, Mango247
Let $a ,\; b,\; c$ be positive real numbers such that
$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1.$
Prove that
$b^2+3 \ge 2b(a+c).$
Z K Y
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sqing
41890 posts
#2 • 4 Y
Y by a_friendwr_a, Ha_ha_ha, Mango247, Mango247
can_hang2007 wrote:
Let $a ,\; b,\; c$ be positive real numbers such that $\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1.$ Prove that
$$b^2+3 \ge 2b(a+c).$$
Good.
Z K Y
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ZHEKSHEN
25 posts
#3 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
By Titu: 3>= (a^2) + (b^2) + (c^2)
Now we can rewrite inequality:
2(b^2)+ (a^2) + (c^2)>= 2ab+ 2bc
By AM-GM:
(b^2)+(a^2)>= 2ab
(b^2)+(c^2)>= 2bc
Summing up two inequality, we have done
P.S: sorry for latex
Z K Y
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menpo
209 posts
#4 • 4 Y
Y by a_friendwr_a, Ha_ha_ha, Mango247, Mango247
ZHEKSHEN wrote:
By Titu: 3>= (a^2) + (b^2) + (c^2)
Now we can rewrite inequality:
2(b^2)+ (a^2) + (c^2)>= 2ab+ 2bc
By AM-GM:
(b^2)+(a^2)>= 2ab
(b^2)+(c^2)>= 2bc
Summing up two inequality, we have done
P.S: sorry for latex

Are you sure that $3\ge a^2+b^2+c^2$ ?
Z K Y
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AleksaS
41 posts
#5 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
Adilet160205 wrote:
ZHEKSHEN wrote:
By Titu: 3>= (a^2) + (b^2) + (c^2)
Now we can rewrite inequality:
2(b^2)+ (a^2) + (c^2)>= 2ab+ 2bc
By AM-GM:
(b^2)+(a^2)>= 2ab
(b^2)+(c^2)>= 2bc
Summing up two inequality, we have done
P.S: sorry for latex

Are you sure that $3\ge a^2+b^2+c^2$ ?
By Titu you get that $9\ge 2(a^2 + b^2 + c^2) + 3$ which means that $3 \ge a^2 + b^2 + c^2$ and by $AM - GM$ you get that $3 \ge 2ab + 2bc - b^2$ from which you get needed inequality.
Z K Y
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pi_quadrat_sechstel
591 posts
#6 • 3 Y
Y by xyzz, a_friendwr_a, Ha_ha_ha
AleksaS wrote:
Adilet160205 wrote:
ZHEKSHEN wrote:
By Titu: 3>= (a^2) + (b^2) + (c^2)
Now we can rewrite inequality:
2(b^2)+ (a^2) + (c^2)>= 2ab+ 2bc
By AM-GM:
(b^2)+(a^2)>= 2ab
(b^2)+(c^2)>= 2bc
Summing up two inequality, we have done
P.S: sorry for latex

Are you sure that $3\ge a^2+b^2+c^2$ ?
By Titu you get that $9\ge 2(a^2 + b^2 + c^2) + 3$ which means that $3 \ge a^2 + b^2 + c^2$ and by $AM - GM$ you get that $3 \ge 2ab + 2bc - b^2$ from which you get needed inequality.

No.

By Titu you get
\[
\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}\geq\frac{9}{2(a^2+b^2+c^2)+3}
\]and not nessecarily
\[
\frac{9}{2(a^2+b^2+c^2)+3}\geq1
\]
Z K Y
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AleksaS
41 posts
#7 • 5 Y
Y by a_friendwr_a, Ha_ha_ha, Mango247, Mango247, Mango247
In this case it would be useful if we can prove that but I don't see a way to do that.
Z K Y
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pi_quadrat_sechstel
591 posts
#8 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
AleksaS wrote:
In this case it would be useful if we can prove that but I don't see a way to do that.

It is not true. For distinct $a,b,c$ with $1=\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}$, we get by Titu
\[
1=\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}>\frac{9}{2(a^2+b^2+c^2)+3}
\]
Z K Y
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alinazarboland
168 posts
#9 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
wrong solution
This post has been edited 1 time. Last edited by alinazarboland, Jan 8, 2021, 10:23 AM
Z K Y
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pi_quadrat_sechstel
591 posts
#10 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
alinazarboland wrote:
can_hang2007 wrote:
Let $a ,\; b,\; c$ be positive real numbers such that
$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1.$
Prove that
$b^2+3 \ge 2b(a+c).$

Let's make $a ,b$ fix. Larger $c$ makes condition stronger and stronger claim .so we can assume that

$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} =1$

now by T2 lemma we have $a^2 + b^2 +c^2 <3$ the rest is easy...

No, we have $a^2+b^2+c^2\geq3$
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arqady
30218 posts
#11 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
alinazarboland wrote:

now by T2 lemma we have $a^2 + b^2 +c^2 <3$ the rest is easy...
Try $c\rightarrow0^+,$ $a=2$ and $b=\sqrt{\frac{\sqrt{89}-7}{4}}.$
This post has been edited 1 time. Last edited by arqady, Jan 8, 2021, 10:10 AM
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csav10
382 posts
#12 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
It is enough to show that
$b^2+3 \ge 2b\sqrt{2(a^2+c^2)},$
which is not hard.
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Universes
96 posts
#13 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
csav10 wrote:
It is enough to show that
$b^2+3 \ge 2b\sqrt{2(a^2+c^2)},$
which is not hard.

Are you sure?
Z K Y
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csav10
382 posts
#14 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
I am sure that $b^2+3 \ge 2b\sqrt{2(a^2+c^2)}$ is wrong, but $\sqrt{2(b^2+1)}\ge b(a+c)$ is true.
This post has been edited 1 time. Last edited by csav10, Jan 11, 2021, 6:30 AM
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pi_quadrat_sechstel
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#15 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
We need to prove
\[
2\left(a^2+b^2+c^2\right)+3\geq(a+b+c)^2+(a-c)^2
\]Thus we can assume w.l.o.g $a\geq b\geq c$. It suffices to prove
\[
\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}<1
\]for reals $a\geq b\geq c>0$ with $3<2b(a+c)-b^2$. Replace $(a,b,c)$ with
\[
\left(a\cdot\sqrt{\frac{3}{2b(a+c)-b^2}},b\cdot\sqrt{\frac{3}{2b(a+c)-b^2}},c\cdot\sqrt{\frac{3}{2b(a+c)-b^2}}\right)
\]and note that $\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}$ rises. It suffices to prove
\[
\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}\leq1
\]for reals $a\geq b\geq c>0$ with $3=2b(a+c)-b^2$. Write the inequality in the homogenous from
\[
\frac{1}{3c^2+2b^2+2b(a+c)}+\frac{1}{3a^2+2b^2+2b(a+c)}+\frac{1}{3a^2+3c^2+2b(a+c)-b^2}\leq\frac{1}{2b(a+c)-b^2}
\]Define $x,y\geq0$ by $b=c+x,a=b+y$. The inequality is eqiuvalent to
\begin{align*}
36c^4(x-y)^2+6c^3x(x-y)^2+54c^4x^2+54c^4y^2+274c^3x^3+132c^3xy^2+80c^3y^3\\
+330c^2x^4+228c^2x^3y+42c^2x^2y^2+72c^2xy^3+30c^2y^4+180cx^5+255cxy^4\\
+84cx^3y^2+12cx^2y^3+6cxy^4+40x^6+90x^5y+63x^4y^2+18x^3y^3+3x^2y^4\geq0
\end{align*}which is obviously true.
This post has been edited 1 time. Last edited by pi_quadrat_sechstel, Jan 8, 2021, 9:45 PM
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Universes
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#16 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
You can't order the variables.
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pi_quadrat_sechstel
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#17 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
Universes wrote:
You can't order the variables.

I can. Since $b^2+3 \ge 2b(a+c)\Leftrightarrow2\left(a^2+b^2+c^2\right)+3\geq(a+b+c)^2+(a-c)^2$ the $RHS$ is maximal for $a\geq b\geq c$.
This post has been edited 1 time. Last edited by pi_quadrat_sechstel, Jan 9, 2021, 9:49 AM
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#18 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
can_hang2007 wrote:
Let $a ,\; b,\; c$ be positive real numbers such that
$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1.$
Prove that
$b^2+3 \ge 2b(a+c).$

Prove that $\lambda=1$ is the best constant with
\[
2(a^2+b^2+c^2)+3\geq(a+b+c)^2+\lambda(a-c)^2
\]for all nonnegative reals $a,b,c$ with
\[
\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1
\]
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#19 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
pi_quadrat_sechstel wrote:
Universes wrote:
You can't order the variables.

I can. Since $b^2+3 \ge 2b(a+c)\Leftrightarrow2\left(a^2+b^2+c^2\right)+3\geq(a+b+c)^2+(a-c)^2$ the $LHS$ is maximal for $a\geq b\geq c$.

LHS or RHS?
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pi_quadrat_sechstel
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#21 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
RHS. I've corrected it now.
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#22 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
pi_quadrat_sechstel wrote:
pi_quadrat_sechstel wrote:
Universes wrote:
You can't order the variables.

I can. Since $b^2+3 \ge 2b(a+c)\Leftrightarrow2\left(a^2+b^2+c^2\right)+3\geq(a+b+c)^2+(a-c)^2$ the $RHS$ is maximal for $a\geq b\geq c$.

So , also you have to consider the case $c\ge b\ge a. $
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#25 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
Universes wrote:

So , also you have to consider the case $c\ge b\ge a. $

No, we can switch $a,c$.
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#26 • 3 Y
Y by a_friendwr_a, Ha_ha_ha, Mango247
Your proof is equivalent to knowing the condition $b^2+3 \ge 2b (a+c) $ and proving the inequality \[
\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}\leq1
\].
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pi_quadrat_sechstel
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#27 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
Universes wrote:
Your proof is equivalent to knowing the condition $b^2+3 \ge 2b (a+c) $ and proving the inequality \[
\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}\leq1
\].

It proves $b^2+3<2b(a+c)\Rightarrow\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}<1$. Since $(\lnot B\Rightarrow\lnot A)\Leftrightarrow(A\Rightarrow B)$ it proves $\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}\geq1\Rightarrow b^2+3\geq2b(a+c)$, too.
This post has been edited 1 time. Last edited by pi_quadrat_sechstel, Jan 9, 2021, 10:19 AM
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csav10
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#28 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
The equality occurs for $a=b=c=1$.
This post has been edited 1 time. Last edited by csav10, Jan 11, 2021, 6:32 AM
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math90
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#29 • 7 Y
Y by arqady, a_friendwr_a, xyzz, hangb6pbc, dragonheart6, mudok, Ha_ha_ha
can_hang2007 wrote:
Let $a ,\; b,\; c$ be positive real numbers such that
$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1.$
Prove that
$b^2+3 \ge 2b(a+c).$

After expanding, the condition is equivalent to
\[2\left(a^2+b^2+c^2+1\right)\ge\prod\left(a^2+b^2\right)\]Details

Equivalently, we have
\[2+\frac{2\left(b^2+1\right)}{a^2+c^2}\ge\left(a^2+b^2\right)\left(b^2+c^2\right)\]Since $2\left(a^2+c^2\right)\ge \left(a+c\right)^2$, we have
\[2+\frac{4\left(b^2+1\right)}{\left(a+c\right)^2}\ge 2+\frac{2\left(b^2+1\right)}{a^2+c^2}\]Moreover, by C-S inequality we have
\[\left(a^2+b^2\right)\left(b^2+c^2\right)\ge b^2\left(a+c\right)^2\]Hence,
\begin{align*}
&2+\frac{4\left(b^2+1\right)}{\left(a+c\right)^2}\ge b^2\left(a+c\right)^2\\
&\implies 2\left(a+c\right)^2+4\ge b^2\left(a+c\right)^4-4b^2\\
&\implies 2\left[\left(a+c\right)^2+2\right]\ge b^2\left[\left(a+c\right)^2-2\right]\left[\left(a+c\right)^2+2\right]\\
&\implies 2\ge b^2\left[\left(a+c\right)^2-2\right]\\
&\implies b^2\left(a+c\right)^2\le 2\left(b^2+1\right)\le\frac{\left(b^2+1+2\right)^2}{4}=\frac{\left(b^2+3\right)^2}{4}\\
&\implies 2b\left(a+c\right)\le b^2+3
\end{align*}as desired.
This post has been edited 1 time. Last edited by math90, Jul 30, 2021, 12:10 PM
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arqady
30218 posts
#30 • 3 Y
Y by a_friendwr_a, math90, Ha_ha_ha
Beautiful proof! :-D
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csav10
382 posts
#31 • 1 Y
Y by Ha_ha_ha
arqady wrote:
Beautiful proof! :-D
I agree with you. :flex:
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pi_quadrat_sechstel
591 posts
#32 • 2 Y
Y by Ha_ha_ha, Mango247
pi_quadrat_sechstel wrote:
can_hang2007 wrote:
Let $a ,\; b,\; c$ be positive real numbers such that
$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1.$
Prove that
$b^2+3 \ge 2b(a+c).$

Prove that $\lambda=1$ is the best constant with
\[
2(a^2+b^2+c^2)+3\geq(a+b+c)^2+\lambda(a-c)^2
\]for all nonnegative reals $a,b,c$ with
\[
\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1
\]

Bump.
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mudok
3377 posts
#33 • 3 Y
Y by dragonheart6, Ha_ha_ha, xyzz
can_hang2007 wrote:
Let $a ,\; b,\; c$ be positive real numbers such that
$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1.$
Prove that
$b^2+3 \ge 2b(a+c).$

$\sum_{cyc}\frac{1}{a^2+b^2+1} \ge 1 \iff 4\ge \sum_{cyc} \frac{(a+b)^2}{a^2+b^2+1} + \sum_{cyc} \frac{(a-b)^2}{a^2+b^2+1} $

By Cauchy-Schwarz:

$ \sum_{cyc} \frac{(a+b)^2}{a^2+b^2+1}\ge  \frac{4(a+b+c)^2}{2(a^2+b^2+c^2)+3}$

$ \sum_{cyc} \frac{(a-b)^2}{a^2+b^2+1}\ge \frac{4(a-c)^2}{2(a^2+b^2+c^2)+3}$

So, we have:

$4\ge \frac{4(a+b+c)^2+4(a-c)^2}{2(a^2+b^2+c^2)+3} \iff b^2+3 \ge 2b(a+c).$
This post has been edited 4 times. Last edited by mudok, Jan 11, 2021, 5:59 AM
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sqing
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#34 • 1 Y
Y by Ha_ha_ha
Beautiful .
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csav10
382 posts
#35 • 2 Y
Y by Ha_ha_ha, Mango247
Actually, the stronger inequality holds:
$$\sqrt{2(b^2+1)}\ge b(a+c).$$
pi_quadrat_sechstel wrote:

Prove that $\lambda=1$ is the best constant with
\[
2(a^2+b^2+c^2)+3\geq(a+b+c)^2+\lambda(a-c)^2
\]for all nonnegative reals $a,b,c$ with
\[
\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1
\]
The hypothesis
$$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1$$is satisfied for $b=c=0$ and all $a\ge 0$, when the inequality
$$2(a^2+b^2+c^2)+3\geq(a+b+c)^2+\lambda(a-c)^2$$becomes
$$3\ge( \lambda -1)a^2.$$Clearly, this is true for any $a\ge 0$ if and only if $\lambda\le 1$.
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mudok
3377 posts
#36 • 3 Y
Y by Ha_ha_ha, Mango247, Mango247
csav10 wrote:
Actually, the stronger inequality holds:
$$\sqrt{2(b^2+1)}\ge b(a+c).$$

You can see its proof in math90's proof. See post #29
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sqing
41890 posts
#37 • 1 Y
Y by Ha_ha_ha
can_hang2007 wrote:
Let $a ,\; b,\; c$ be positive real numbers such that $\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1.$ Prove that
$$b^2+3 \ge 2b(a+c).$$
Solution of Zhangyanzong:
$$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1\iff  2\geq \frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1}+\frac{a^2+b^2}{a^2+b^2+1} $$$$\frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1}+\frac{a^2+b^2}{a^2+b^2+1} \geq \frac{(\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+\sqrt{a^2+b^2})^2}{2(a^2+b^2+c^2)+3}$$$$2\geq \frac{(\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+\sqrt{a^2+b^2})^2}{2(a^2+b^2+c^2)+3}$$$$b^2+3\geq \sqrt{(a^2+b^2)(a^2+c^2)}+\sqrt{(c^2+a^2)(c^2+b^2)}+\sqrt{(a^2+b^2)(b^2+c^2)}-a^2-c^2$$$$\geq a^2+bc+c^2+ab+ab+bc-a^2-c^2=2b(a+c).$$
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sqing
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#38
Y by
can_hang2007 wrote:
Let $a ,\; b,\; c$ be positive real numbers such that $\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1.$ Prove that
$$b^2+3 \ge 2b(a+c).$$
Maybe inspired by here

Suppose that $ a$,$ b$,$ c$ be three positive real numbers such that $ a+b+c=3$ . Prove that $$ \frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leq \frac{3}{4}$$Iran TST 2009-Day1-P3
This post has been edited 3 times. Last edited by sqing, Jan 15, 2021, 2:06 PM
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#39 • 3 Y
Y by Mango247, Mango247, Mango247
Let $a,b,c$ be positive real numbers such $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1} \geq 1$.Prove that$$a^2b^2+b^2c^2+c^2a^2 \leq 3.$$Let $a,b,c$ be positive real numbers such $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1} \geq 1$.Prove that$$ab+bc+ca \leq 3.$$Solution:
$$\frac{2+c^2}{(a+b+c)^2}\geq \frac{1+1+c^2}{(a^2+b^2+1)(1+1+c^2)}=\frac{1}{(a^2+b^2+1)}$$$$6+a^2+b^2+c^2\geq (a+b+c)^2$$$$ab+bc+ca \leq 3$$h
This post has been edited 2 times. Last edited by sqing, Nov 5, 2021, 10:24 AM
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#40
Y by
Let $a ,b,c$ be positive real numbers such that $\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}\geq 1.$ Prove that
$$b^2+3 \ge b\sqrt {2(a^2+c^2)}$$
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sqing
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#41
Y by
Let $a,b,c$ be positive real numbers such that $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}\geq 1.$ Prove that$$c^k+\frac{3}{c^k}\geq 2(a^k+b^k)$$Where $k=1,2.$
Let $a,b,c$ be positive real numbers such that $\frac{a+b}{a^2+b^2+1}+\frac{b+c}{b^2+c^2+1}+\frac{c+a}{c^2+a^2+1}\geq 2.$ Prove that
$$c^k+\frac{3}{c^k}\geq 2(a^k+b^k)$$Where $k=1,2,3,4.$
Let $a,b,c$ be positive real numbers such that $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}\leq 1.$ Prove that$$c+\frac{3}{c}\geq 2(a+b)$$
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