Happy Thanksgiving! Please note that there are no classes November 25th-December 1st.

G
Topic
First Poster
Last Poster
k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile!

Whether you've taken our classes at AoPS Online or AoPS Academies or reached the top echelons of our competition training with our Worldwide Online Olympiad Training (WOOT) program, you can now add your AoPS experience to the education section on your LinkedIn profile.

Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1
Sunday, Nov 3 - Mar 9
Tuesday, Nov 5 - Mar 11
Friday, Dec 6 - Apr 4
Sunday, Jan 5 - Apr 20
Wednesday, Jan 15 - Apr 30
Monday, Feb 3 - May 19
Sunday, Mar 2 - Jun 22
Friday, Mar 28 - Jul 18
Sunday, Apr 13 - Aug 10

Prealgebra 1 Self-Paced

Prealgebra 2
Thursday, Nov 7 - Mar 13
Monday, Dec 2 - Mar 31
Wednesday, Jan 8 - Apr 23
Sunday, Jan 19 - May 4 (1:00 - 2:15 pm ET/10:00 - 11:15 am PT)
Monday, Jan 27 - May 12
Tuesday, Jan 28 - May 13 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT)
Sunday, Feb 16 - Jun 8
Tuesday, Mar 25 - Jul 8
Sunday, Apr 13 - Aug 10

Prealgebra 2 Self-Paced

Introduction to Algebra A
Friday, Nov 8 - Mar 14
Wednesday, Dec 11 - Apr 9
Tuesday, Jan 7 - Apr 22
Wednesday, Jan 29 - May 14
Sunday, Feb 16 - Jun 8 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)
Sunday, Mar 23 - Jul 20
Monday, Apr 7 - Jul 28

Introduction to Algebra A Self-Paced

Introduction to Counting & Probability
Thursday, Dec 5 - Mar 6
Wednesday, Jan 8 - Mar 26
Thursday, Jan 30 - Apr 17
Sunday, Feb 9 - Apr 27 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)
Sunday, Mar 16 - Jun 8
Wednesday, Apr 16 - Jul 2

Introduction to Counting & Probability Self-Paced

Introduction to Number Theory
Monday, Dec 2 - Mar 3
Tuesday, Jan 28 - Apr 15
Sunday, Feb 16 - May 4
Monday, Mar 17 - Jun 9
Thursday, Apr 17 - Jul 3

Introduction to Algebra B
Wednesday, Dec 11 - Apr 9
Tuesday, Jan 28 - May 13
Thursday, Feb 13 - May 29
Sunday, Mar 2 - Jun 22
Monday, Mar 17 - Jul 7
Wednesday, Apr 16 - Jul 30

Introduction to Geometry
Monday, Nov 11 - May 12
Wednesday, Nov 13 - May 14 (9:30 - 11:00 pm ET/6:30 - 8:00 pm PT)
Tuesday, Dec 10 - Jun 3
Wednesday, Jan 8 - Jun 18
Thursday, Jan 30 - Jul 10
Friday, Feb 14 - Aug 1
Tuesday, Mar 4 - Aug 12
Sunday, Mar 23 - Sep 21
Wednesday, Apr 23 - Oct 1

Paradoxes and Infinity
Sat & Sun, Nov 16 - Nov 17 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Nov 10 - May 11
Tuesday, Dec 3 - May 27
Friday, Jan 17 - Jun 27
Wednesday, Feb 12 - Jul 23
Sunday, Mar 16 - Sep 14
Tuesday, Mar 25 - Sep 2
Monday, Apr 21 - Oct 13

Intermediate Counting & Probability
Thursday, Nov 7 - Mar 27
Monday, Feb 10 - Jun 16
Sunday, Mar 23 - Aug 3

Intermediate Number Theory
Thursday, Feb 20 - May 8
Friday, Apr 11 - Jun 27

Precalculus
Sunday, Nov 10 - Apr 27
Tuesday, Dec 10 - May 20
Wednesday, Jan 8 - Jun 4
Tuesday, Feb 25 - Jul 22
Sunday, Mar 16 - Aug 24
Wednesday, Apr 9 - Sep 3

Advanced: Grades 9-12

Olympiad Geometry
Wednesday, Mar 5 - May 21

Calculus
Tuesday, Dec 10 - Jun 10
Friday, Feb 28 - Aug 22
Sunday, Mar 30 - Oct 5

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Mon, Wed & Fri, Dec 2 - Jan 10 (meets three times each week!)
Tuesday, Feb 4 - Apr 22
Sunday, Mar 23 - Jun 15
Wednesday, Apr 16 - Jul 2

MATHCOUNTS/AMC 8 Advanced
Tuesday, Nov 5 - Feb 11
Mon, Wed & Fri, Dec 2 - Jan 10 (meets three times each week!)
Tue, Thurs & Sun, Dec 10 - Jan 19 (meets three times each week!)
Sunday, Feb 16 - May 4
Friday, Apr 11 - Jun 27

Special AMC 8 Problem Seminar A
Sat & Sun, Jan 11 - Jan 12 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)

Special AMC 8 Problem Seminar B
Sat & Sun, Jan 18 - Jan 19 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)

AMC 10 Problem Series
Sunday, Feb 9 - Apr 27
Tuesday, Mar 4 - May 20
Monday, Mar 31 - Jun 23

AMC 10 Final Fives
Sunday, Feb 9 - Mar 2 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)

Special AMC 10 Problem Seminar B
Sat & Sun, Nov 2 - Nov 3 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)

AMC 12 Problem Series
Sunday, Feb 23 - May 11

AMC 12 Final Fives
Sunday, Feb 9 - Mar 2 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)

Special AMC 12 Problem Seminar B
Sat & Sun, Nov 2 - Nov 3 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)

AIME Problem Series A
Tue, Thurs & Sun, Jan 7 - Feb (meets three times each week!)

AIME Problem Series B
Mon, Wed & Fri, Jan 6 - Jan 31 (meets three times each week!)

Special AIME Problem Seminar A
Sat & Sun, Jan 25 - Jan 26 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)

Special AIME Problem Seminar B
Sat & Sun, Feb 1 - Feb 2 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)

F=ma Problem Series
Wednesday, Feb 19 - May 7

Programming

Introduction to Programming with Python
Monday, Dec 2 - Mar 3
Friday, Jan 17 - Apr 4
Sunday, Feb 16 - May 4
Monday, Mar 24 - Jun 16

Intermediate Programming with Python
Tuesday, Feb 25 - May 13

Science

Introduction to Physics
Tuesday, Dec 10 - Mar 11
Friday, Feb 7 - Apr 25
Sunday, Mar 30 - Jun 22

Physics 1: Mechanics
Sunday, Feb 9 - Aug 3
Tuesday, Mar 25 - Sep 2

Relativity
Sat & Sun, Dec 14 - Dec 15 (4:00 - 7:00 pm ET/1:00 - 4:00pm PT)
0 replies
jlacosta
Nov 1, 2024
0 replies
a+b+c+abc=4 with two equality cases
KhuongTrang   49
N 2 minutes ago by arqady
Source: own
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c+abc=4.$ Prove that
$$\color{blue}{\sqrt{14a+b+c} +\sqrt{14b+a+c} +\sqrt{14c+b+a}\le 2+2\sqrt{5}\cdot\sqrt{a+b+c+2}. }$$Equality holds iff $a=b=c=1$ or $a=b=2,c=0$ and any permutations.
49 replies
KhuongTrang
Mar 11, 2024
arqady
2 minutes ago
Symmetric geo with tangent circles
AlephG_64   1
N 5 minutes ago by sami1618
Source: 2nd AGO P4
Let $ABC$ be a scalene triangle. The perpendicular bisector of $BC$ intersects lines $AB$ and $AC$ at $A_b$ and $A_c$ respectively. Let $O_a$ denote the circumcenter of triangle $AA_bA_c$. Define $O_b, O_c$ similarly.
Prove that the circumcircle of triangle $O_aO_bO_c$ is tangent to the circumcircle of triangle $ABC$.

Proposed by Atavic
1 reply
AlephG_64
2 hours ago
sami1618
5 minutes ago
a^2+ b^2+ c^2
sqing   3
N 37 minutes ago by sqing
Source: Own
Let $ a,b, c$ be reals such that $a +b =2 , b^2+2c^2= 9 $ and $ 3c^2+4a^2 =48.$ Prove that$$a^2+ b^2+ c^2=14$$Let $ a,b ,c$ be reals such that $a +b =1 , b^2+2c^2= 9 $ and $ 3c^2+4a^2 =48.$ Prove that$$a^2+ b^2+ c^2=\frac{7(101-6\sqrt{41})}{25}$$
3 replies
sqing
Yesterday at 1:48 AM
sqing
37 minutes ago
Needed help on inequalities
fAaAtDoOoG   10
N an hour ago by arqady
Hello guys, this is my first post. I've encountered an inequality and struggled to solve it. If anyone can solve it, that would be awesome. Any help would be greatly appreciated!!!

$$\frac{b+c}{\sqrt{a^{2}+bc}} + \frac{c+a}{\sqrt{b^{2}+ac}} + \frac{a+b}{\sqrt{c^{2}+ab}} > 4, a,b,c \in \mathbb{R}^{+}$$
10 replies
fAaAtDoOoG
Yesterday at 1:28 AM
arqady
an hour ago
No more topics!
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi_quadrat_sechstel
581 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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi_quadrat_sechstel
581 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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
alinazarboland
163 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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi_quadrat_sechstel
581 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 $a^2+b^2+c^2\geq3$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
arqady
29868 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi_quadrat_sechstel
581 posts
#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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Universes
96 posts
#16 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
You can't order the variables.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi_quadrat_sechstel
581 posts
#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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi_quadrat_sechstel
581 posts
#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
\[
\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1
\]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Universes
96 posts
#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?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi_quadrat_sechstel
581 posts
#21 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
RHS. I've corrected it now.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Universes
96 posts
#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. $
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi_quadrat_sechstel
581 posts
#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 $a,c$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Universes
96 posts
#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 \[
\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}\leq1
\].
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi_quadrat_sechstel
581 posts
#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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
csav10
382 posts
#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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
math90
1449 posts
#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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
arqady
29868 posts
#30 • 3 Y
Y by a_friendwr_a, math90, Ha_ha_ha
Beautiful proof! :-D
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
csav10
382 posts
#31 • 1 Y
Y by Ha_ha_ha
arqady wrote:
Beautiful proof! :-D
I agree with you. :flex:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pi_quadrat_sechstel
581 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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mudok
3373 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#34 • 1 Y
Y by Ha_ha_ha
Beautiful .
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $\lambda\le 1$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mudok
3373 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 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:
$$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1\iff  2\geq \frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1}+\frac{a^2+b^2}{a^2+b^2+1} $$$$\frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1}+\frac{a^2+b^2}{a^2+b^2+1} \geq \frac{(\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+\sqrt{a^2+b^2})^2}{2(a^2+b^2+c^2)+3}$$$$2\geq \frac{(\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+\sqrt{a^2+b^2})^2}{2(a^2+b^2+c^2)+3}$$$$b^2+3\geq \sqrt{(a^2+b^2)(a^2+c^2)}+\sqrt{(c^2+a^2)(c^2+b^2)}+\sqrt{(a^2+b^2)(b^2+c^2)}-a^2-c^2$$$$\geq a^2+bc+c^2+ab+ab+bc-a^2-c^2=2b(a+c).$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#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
$$b^2+3 \ge b\sqrt {2(a^2+c^2)}$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sqing
39045 posts
#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$$c+\frac{3}{c}\geq 2(a+b)$$
Attachments:
This post has been edited 1 time. Last edited by sqing, Nov 13, 2021, 9:09 AM
Z K Y
N Quick Reply
G
H
=
a