A simple inequality for highschool students

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can_hang2007

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can_hang2007

#1 h Nov 3, 2010, 12:00 PM • 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
$b^2+3 \ge 2b(a+c).$

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sqing

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sqing

#2 h Jan 7, 2021, 12:38 PM • 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

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ZHEKSHEN

#3 h Jan 7, 2021, 2:28 PM • 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

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menpo

#4 h Jan 7, 2021, 2:40 PM • 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

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AleksaS

#5 h Jan 7, 2021, 9:06 PM • 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

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pi_quadrat_sechstel

#6 h Jan 7, 2021, 10:30 PM • 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

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AleksaS

#7 h Jan 8, 2021, 12:20 AM • 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

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pi_quadrat_sechstel

#8 h Jan 8, 2021, 8:27 AM • 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

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alinazarboland

#9 h Jan 8, 2021, 9:38 AM • 2 Y

Y by a_friendwr_a, Ha_ha_ha

wrong solution

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pi_quadrat_sechstel

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pi_quadrat_sechstel

#10 h Jan 8, 2021, 10:07 AM • 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

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arqady

#11 h Jan 8, 2021, 10:10 AM • 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 $b=\sqrt{\frac{\sqrt{89}-7}{4}}.$

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csav10

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csav10

#12 h Jan 8, 2021, 3:25 PM • 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

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Universes

#13 h Jan 8, 2021, 3:38 PM • 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?

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csav10

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csav10

#14 h Jan 8, 2021, 5:51 PM • 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.

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pi_quadrat_sechstel

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pi_quadrat_sechstel

#15 h Jan 8, 2021, 8:50 PM • 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.

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Universes

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Universes

#16 h Jan 9, 2021, 6:43 AM • 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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pi_quadrat_sechstel

#17 h Jan 9, 2021, 8:47 AM • 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 $a\geq b\geq c$ .

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pi_quadrat_sechstel

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pi_quadrat_sechstel

#18 h Jan 9, 2021, 9:16 AM • 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

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Universes

#19 h Jan 9, 2021, 9:47 AM • 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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pi_quadrat_sechstel

#21 h Jan 9, 2021, 9:49 AM • 2 Y

Y by a_friendwr_a, Ha_ha_ha

RHS. I've corrected it now.

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Universes

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Universes

#22 h Jan 9, 2021, 9:51 AM • 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 $RHS$ 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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pi_quadrat_sechstel

#25 h Jan 9, 2021, 9:58 AM • 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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Universes

#26 h Jan 9, 2021, 10:04 AM • 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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pi_quadrat_sechstel

#27 h Jan 9, 2021, 10:17 AM • 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.

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csav10

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csav10

#28 h Jan 9, 2021, 6:38 PM • 2 Y

Y by a_friendwr_a, Ha_ha_ha

The equality occurs for $a=b=c=1$ .

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math90

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math90

#29 h Jan 10, 2021, 2:16 PM • 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

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arqady

#30 h Jan 10, 2021, 2:54 PM • 3 Y

Y by a_friendwr_a, math90, Ha_ha_ha

Beautiful proof! :-D

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csav10

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csav10

#31 h Jan 10, 2021, 9:08 PM • 1 Y

Y by Ha_ha_ha

arqady wrote:

Beautiful proof! :-D

I agree with you. :flex:

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pi_quadrat_sechstel

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pi_quadrat_sechstel

#32 h Jan 10, 2021, 9:35 PM • 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

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mudok

#33 h Jan 11, 2021, 4:52 AM • 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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sqing

#34 h Jan 11, 2021, 6:37 AM • 1 Y

Y by Ha_ha_ha

Beautiful .

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csav10

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csav10

#35 h Jan 11, 2021, 7:06 AM • 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

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mudok

#36 h Jan 11, 2021, 8:33 AM • 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

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sqing

#37 h Jan 15, 2021, 1:29 PM • 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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#38 h Jan 15, 2021, 1:52 PM

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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

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#39 h Jan 20, 2021, 1:19 PM • 3 Y

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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

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#40 h Nov 7, 2021, 12:58 PM

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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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#41 h Nov 12, 2021, 12:34 AM

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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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