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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]
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[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]
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[*]Did you run out of time?[/list][/list]
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[*]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]
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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 38 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
38 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
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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
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.
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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
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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 $3\ge a^2+b^2+c^2$ ?
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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.
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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
\[
\frac{9}{2(a^2+b^2+c^2)+3}\geq1
\]
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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.
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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
\[
1=\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1}>\frac{9}{2(a^2+b^2+c^2)+3}
\]
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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
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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 $a^2+b^2+c^2\geq3$
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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
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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
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
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Universes
96 posts
#16 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
You can't order the variables.
Z K Y
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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
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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
\[
\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1
\]
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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?
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pi_quadrat_sechstel
581 posts
#21 • 2 Y
Y by a_friendwr_a, Ha_ha_ha
RHS. I've corrected it now.
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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. $
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pi_quadrat_sechstel
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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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Universes
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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
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
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arqady
29868 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
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.
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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
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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
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
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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:
$$\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}+\frac{1}{a^2+b^2+1} \ge 1\iff  2\geq \frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1}+\frac{a^2+b^2}{a^2+b^2+1} $$$$\frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1}+\frac{a^2+b^2}{a^2+b^2+1} \geq \frac{(\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+\sqrt{a^2+b^2})^2}{2(a^2+b^2+c^2)+3}$$$$2\geq \frac{(\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+\sqrt{a^2+b^2})^2}{2(a^2+b^2+c^2)+3}$$$$b^2+3\geq \sqrt{(a^2+b^2)(a^2+c^2)}+\sqrt{(c^2+a^2)(c^2+b^2)}+\sqrt{(a^2+b^2)(b^2+c^2)}-a^2-c^2$$$$\geq a^2+bc+c^2+ab+ab+bc-a^2-c^2=2b(a+c).$$
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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
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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
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sqing
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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
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$$c+\frac{3}{c}\geq 2(a+b)$$
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