Inequality with \sqrt{a}+\sqrt{b}+\sqrt{c}

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Inequality with \sqrt{a}+\sqrt{b}+\sqrt{c}

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Source: Stars of Mathematics 2015 Junior Level #1

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#1 h Jan 2, 2016, 11:16 AM • 1 Y

Y by Adventure10

Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.

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#2 h Jan 2, 2016, 11:39 AM • 3 Y

Y by Misha57, mijail, Adventure10

A certain substitution of $a=\frac{x}{y+z}$ , $b=\frac{y}{x+z}$ , and $c=\frac{z}{x+y}$ gives you the result.

It now suffices to prove that $\sum_{cyc}\sqrt{\frac{x}{y+z}} \ge 2$ . WLOG $x+y+z=1$ and use $\sqrt{\frac{x}{1-x}} \ge 2x$ .

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#3 h Jan 10, 2016, 4:47 PM • 1 Y

Y by Adventure10

In turn, we could bound the variables in $[0,4]^3$ and see that by Lagrange Multipliers we are done.

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sqing

#4 h Jan 11, 2016, 7:05 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\frac 1{8a^2+1}+\frac 1{8b^2+1}+\frac 1{8c^2+1}\ge 1.$

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mudok

#5 h Jan 14, 2016, 3:44 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\frac 1{8a^2+1}+\frac 1{8b^2+1}+\frac 1{8c^2+1}\ge 1.$

$\iff 4(a^2+b^2+c^2)+1\ge 256^2b^2c^2$ which is obvious.

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sqing

#6 h Jan 15, 2016, 12:52 AM • 2 Y

Y by Adventure10, Mango247

huricane wrote:

Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.

http://www.artofproblemsolving.com/community/c6h1184979p5752974
Let $a,b,c>0$ & $ab+ac+bc+2abc=1$ . Prove, that $\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\leq\frac{3}{2}$

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K.N

#7 h Jun 15, 2016, 10:53 AM • 2 Y

Y by Adventure10, Mango247

huricane wrote:

Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.

By lagrange theorem we easily get that
$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \frac{3\sqrt{2}}{2}$ and the equality occurs when $a=b=c=\frac{1}{2}$

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Reynan

#8 h Jun 15, 2016, 12:02 PM • 1 Y

Y by Adventure10

K.N wrote:

huricane wrote:

Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.

By lagrange theorem we easily get that
$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \frac{3\sqrt{2}}{2}$ and the equality occurs when $a=b=c=\frac{1}{2}$

how about $a=0,b=c=1$ ?

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K.N

#9 h Jun 15, 2016, 12:35 PM • 2 Y

Y by Adventure10, Mango247

Reynan wrote:

K.N wrote:

huricane wrote:

Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.

By lagrange theorem we easily get that
$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \frac{3\sqrt{2}}{2}$ and the equality occurs when $a=b=c=\frac{1}{2}$

how about $a=0,b=c=1$ ?

This works when $a,b,c>0$ those cases must be checked separate

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mudok

#10 h Jun 15, 2016, 12:41 PM • 2 Y

Y by Adventure10, Mango247

K.N wrote:

huricane wrote:

Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.

By lagrange theorem we easily get that
$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \frac{3\sqrt{2}}{2}$ and the equality occurs when $a=b=c=\frac{1}{2}$

Try $c=0.00000000000000001, \ \ a=b$

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#11 h Jun 15, 2016, 2:16 PM • 2 Y

Y by Adventure10, Mango247

huricane wrote:

Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.

$\left[ {Vasile\,\,Cirtoaje} \right]$ :
Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$ .

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arqady

#12 h Jun 15, 2016, 6:10 PM • 3 Y

Y by thinhrost1, Adventure10, Mango247

Nguyenngoctu wrote:

$\left[ {Vasile\,\,Cirtoaje} \right]$ :
Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$ .

I think this inequality was before than Vasile Cirtoaje born.

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#13 h Jun 15, 2016, 11:56 PM • 3 Y

Y by lmht, Adventure10, Mango247

Let $a=\frac{x}{y+z}$ , $b=\frac{y}{x+z}$ , $c=\frac{z}{x+y}$ , we will prove that:

$\sum \sqrt{\frac{x}{y+z}} \ge 2$

If $x=0$ or $y=0$ , $z=0$ it 's very easy to prove by AM-GM.

$\sum \sqrt{\frac{x}{y+z}}= \sum \frac{x}{\sqrt{x(y+z)}}\ge \sum \frac{2x}{x+y+z}=2$ ( By AM-GM: $2\sqrt{x(y+z)} \le x+y+z$ )

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K.N

#14 h Jun 16, 2016, 4:14 AM • 2 Y

Y by Adventure10, Mango247

K.N wrote:

huricane wrote:

Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.

By lagrange theorem we easily get that
$\sqrt{a}+\sqrt{b}+\sqrt{c}\ge \frac{3\sqrt{2}}{2}$ and the equality occurs when $a=b=c=\frac{1}{2}$

I can't get why my solution is not correct
Can some one explain for me?!
What should we do if we want to solve it by Lagrange method?!

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mudok

#16 h Jun 16, 2016, 5:58 AM • 2 Y

Y by Adventure10, Mango247

K.N wrote:

I can't get why my solution is not correct
Can some one explain for me?!

Try $a=b=\frac{15}{16}, \ \ c=\frac{1}{30}$ . Are you still going to prove wrong inequality ?

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Orkhan-Ashraf_2002

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Orkhan-Ashraf_2002

#17 h Jun 16, 2016, 6:02 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

Nguyenngoctu wrote:

$\left[ {Vasile\,\,Cirtoaje} \right]$ :
Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$ .

I think this inequality was before than Vasile Cirtoaje born.

We have $a\not=0,b\not=0,c\not=0$ .
We'll show that $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c}$ ,for every $a,b,c$ nonegative numbers.
We have $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c} \Longleftrightarrow \frac{a}{b+c}\ge (\frac{2a}{a+b+c})^2 \Longleftrightarrow (a+b+c)^2\ge 4a(b+c) \Longleftrightarrow (b+c-a)^2\ge 0$ with equality iff $a=b+c$ .
Now we easily obtain $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge \frac{2(a+b+c)}{a+b+c}=2$ with equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a+b+c=0$ $\Longrightarrow$ $a=b=c=0$ $which$ $is$ $impossible.$ $\Longrightarrow$ . $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}$ > $2$

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K.N

#19 h Jun 16, 2016, 7:56 AM • 2 Y

Y by Adventure10, Mango247

mudok wrote:

K.N wrote:

I can't get why my solution is not correct
Can some one explain for me?!

Try $a=b=\frac{15}{16}, \ \ c=\frac{1}{30}$ . Are you still going to prove wrong inequality ?

I know what you say
But i mean if you have a correct solution by Lagrange method i'll be satisfied if you tell me
Thanks:)

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#20 h Jun 16, 2016, 8:54 AM • 1 Y

Y by Adventure10

Orkhan-Ashraf_2002 wrote:

arqady wrote:

Nguyenngoctu wrote:

$\left[ {Vasile\,\,Cirtoaje} \right]$ :
Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$ .

I think this inequality was before than Vasile Cirtoaje born.

We'll show that $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c}$ ,for every $a,b,c$ nonegative numbers.
We have $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c} \Longleftrightarrow \frac{a}{b+c}\ge (\frac{2a}{a+b+c})^2 \Longleftrightarrow (a+b+c)^2\ge 4a(b+c) \Longleftrightarrow (b+c-a)^2\ge 0$ with equality iff $a=b+c$ .
Now we easily obtain $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge \frac{2(a+b+c)}{a+b+c}=2$ with equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a+b+c=0$ $\Longrightarrow$ $a=b=c=0$ .

$a=b=c=0$ then $0>2$

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Orkhan-Ashraf_2002

#21 h Jun 16, 2016, 1:33 PM • 2 Y

Y by Adventure10, Mango247

thinhrost1 wrote:

Let $a=\frac{x}{y+z}$ , $b=\frac{y}{x+z}$ , $c=\frac{z}{x+y}$ , we will prove that:

$\sum \sqrt{\frac{x}{y+z}} \ge 2$

If $x=0$ or $y=0$ , $z=0$ it 's very easy to prove by AM-GM.

$\sum \sqrt{\frac{x}{y+z}}= \sum \frac{x}{\sqrt{x(y+z)}}\ge \sum \frac{2x}{x+y+z}=2$ ( By AM-GM: $2\sqrt{x(y+z)} \le x+y+z$ )

EQUALITY?

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Orkhan-Ashraf_2002

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Orkhan-Ashraf_2002

#23 h Jun 16, 2016, 1:43 PM • 1 Y

Y by Adventure10

Nguyenngoctu wrote:

huricane wrote:

Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.

$\left[ {Vasile\,\,Cirtoaje} \right]$ :
Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$ .

This inequality wrong.Right inequality $\Longrightarrow$ $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2$

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#25 h Jun 16, 2016, 2:13 PM • 1 Y

Y by Adventure10

Orkhan-Ashraf_2002 wrote:

This inequality wrong.Right inequality $\Longrightarrow$ $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2$

Hi Orxan,then you proof is right? I think your proof is false.

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#26 h Jun 16, 2016, 2:15 PM • 2 Y

Y by lmht, Adventure10

Orkhan-Ashraf_2002 wrote:

thinhrost1 wrote:

Let $a=\frac{x}{y+z}$ , $b=\frac{y}{x+z}$ , $c=\frac{z}{x+y}$ , we will prove that:

$\sum \sqrt{\frac{x}{y+z}} \ge 2$

If $x=0$ or $y=0$ , $z=0$ it 's very easy to prove by AM-GM.

$\sum \sqrt{\frac{x}{y+z}}= \sum \frac{x}{\sqrt{x(y+z)}}\ge \sum \frac{2x}{x+y+z}=2$ ( By AM-GM: $2\sqrt{x(y+z)} \le x+y+z$ )

EQUALITY?

In the first case: equality when $x=0, y=z$

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Orkhan-Ashraf_2002

#27 h Jun 16, 2016, 3:07 PM • 2 Y

Y by Adventure10, Mango247

thinhrost1 wrote:

Orkhan-Ashraf_2002 wrote:

thinhrost1 wrote:

Let $a=\frac{x}{y+z}$ , $b=\frac{y}{x+z}$ , $c=\frac{z}{x+y}$ , we will prove that:

$\sum \sqrt{\frac{x}{y+z}} \ge 2$

If $x=0$ or $y=0$ , $z=0$ it 's very easy to prove by AM-GM.

$\sum \sqrt{\frac{x}{y+z}}= \sum \frac{x}{\sqrt{x(y+z)}}\ge \sum \frac{2x}{x+y+z}=2$ ( By AM-GM: $2\sqrt{x(y+z)} \le x+y+z$ )

EQUALITY?

In the first case: equality when $x=0, y=z$

Look my post above.I wrote equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a=b=c=0$ . $Which$ $is$ $impossible$ .Then equality is not a state;Because $a\not=0$ ; $b\not=0$ ; $c\not=0$ .

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#28 h Jun 16, 2016, 3:15 PM • 2 Y

Y by Adventure10, Mango247

Orkhan-Ashraf_2002 wrote:

Look my post above.I wrote equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a=b=c=0$ . $Which$ $is$ $impossible$ .Then equality is not a state;

Orxan,you are false.The equality of $(x,y,z)=(0,y,y),(0,z,z),(x,0,x),(z,0,z),(x,x,0),(y,y,0)$

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Orkhan-Ashraf_2002

#29 h Jun 16, 2016, 3:35 PM • 1 Y

Y by Adventure10

Ferid.---. wrote:

Orkhan-Ashraf_2002 wrote:

Look my post above.I wrote equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a=b=c=0$ . $Which$ $is$ $impossible$ .Then equality is not a state;

Orxan,you are false.The equality of $(x,y,z)=(0,y,y),(0,z,z),(x,0,x),(z,0,z),(x,x,0),(y,y,0)$

$x\not=0;y\not=0;z\not=0.$ If $x=0$ $\Longrightarrow a=0,$ if $y=0 \Longrightarrow b=0,$ if $z=0 \Longrightarrow c=0$ .It is a imbalance.

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#30 h Jun 16, 2016, 3:52 PM • 2 Y

Y by lmht, Adventure10

Orkhan-Ashraf_2002 wrote:

arqady wrote:

Nguyenngoctu wrote:

$\left[ {Vasile\,\,Cirtoaje} \right]$ :
Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$ .

I think this inequality was before than Vasile Cirtoaje born.

We have $a\not=0,b\not=0,c\not=0$ .
We'll show that $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c}$ ,for every $a,b,c$ nonegative numbers.
We have $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c} \Longleftrightarrow \frac{a}{b+c}\ge (\frac{2a}{a+b+c})^2 \Longleftrightarrow (a+b+c)^2\ge 4a(b+c) \Longleftrightarrow (b+c-a)^2\ge 0$ with equality iff $a=b+c$ .
Now we easily obtain $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge \frac{2(a+b+c)}{a+b+c}=2$ with equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a+b+c=0$ $\Longrightarrow$ $a=b=c=0$ $which$ $is$ $impossible.$ $\Longrightarrow$ . $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}$ > $2$

If $a+b+c=0$ then we can't have $\frac{2a}{a+b+c}$ , so we have to seperate into two cases:
+ all of $a,b,c \ne 0$
+or $a \ne 0$ , or $b\ne 0$ , or $c\ne 0$

Cearly in the first case there aren't any a,b,c hold equality

The second equality when $a=0$ , $b=c$ ...

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Orkhan-Ashraf_2002

#31 h Jun 16, 2016, 4:01 PM • 1 Y

Y by Adventure10

Dear thinhrost.
You wrote $a\not=0;b\not=0;c\not=0$ , after you wrote equality when $a=0,b=c.$

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#32 h Jun 16, 2016, 4:02 PM • 1 Y

Y by Adventure10

Orxan your solution is wrong,because is impossible.Right solution is #thinhrost1.

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Orkhan-Ashraf_2002

#33 h Jun 16, 2016, 4:03 PM • 2 Y

Y by Adventure10, Mango247

Ferid.---. wrote:

Orxan your solution is wrong,because is impossible.Right solution is #thinhrost1.

Prove that Farid my solution is wrong.

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#34 h Jun 16, 2016, 5:25 PM • 3 Y

Y by lmht, Adventure10, Mango247

Orkhan-Ashraf_2002 wrote:

Ferid.---. wrote:

Orxan your solution is wrong,because is impossible.Right solution is #thinhrost1.

Prove that Farid my solution is wrong.

It's easy to prove

Orkhan-Ashraf_2002 wrote:

arqady wrote:

Nguyenngoctu wrote:

$\left[ {Vasile\,\,Cirtoaje} \right]$ :
Let a, b, c non-negative, prove that: $\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2$ .

I think this inequality was before than Vasile Cirtoaje born.

We have $a\not=0,b\not=0,c\not=0$ .
We'll show that $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c}$ ,for every $a,b,c$ nonegative numbers.
We have $\sqrt{\frac{a}{b+c}}\ge \frac{2a}{a+b+c} \Longleftrightarrow \frac{a}{b+c}\ge (\frac{2a}{a+b+c})^2 \Longleftrightarrow (a+b+c)^2\ge 4a(b+c) \Longleftrightarrow (b+c-a)^2\ge 0$ with equality iff $a=b+c$ .
Now we easily obtain $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\ge \frac{2(a+b+c)}{a+b+c}=2$ with equality if and only if $a=b+c,b=a+c,c=a+b$ $\Longrightarrow$ $a+b+c=0$ $\Longrightarrow$ $a=b=c=0$ $which$ $is$ $impossible.$ $\Longrightarrow$ . $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}$ > $2$

At here you said: $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}$ > $2$ for all non-negatives $a,b,c$ ( it means: $a,b,c \ge 0$ )

But when a=0, $b=c \ne0$ then $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}=2$

Contradiction

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#35 h Jun 17, 2016, 6:02 AM • 2 Y

Y by Adventure10, Mango247

Yes, i solved this question when $a,b,c>0$ . $\Longrightarrow \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}$ + $\sqrt{\frac{c}{a+b}}$ >2.Sorry,i don't consider $a,b,c\ge 0.$

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sqing

#36 h Aug 2, 2016, 2:46 PM • 1 Y

Y by Adventure10

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $4a+b+c\ge 2.$ (Mathematical Olympiad 2015 A.Petroupolis Hg (9th grade))

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arqady

#38 h Aug 3, 2016, 5:25 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $4a+b+c\ge 2.$

Let $a=\frac{x}{y+z}$ and $b=\frac{y}{x+z}$ , where $x$ , $y$ and $z$ are positives.
Hence, $c=\frac{z}{x+y}$ and by C-S we obtain:
$4a+b+c-2=\frac{4x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}-2\geq\frac{(2x+y+z)^2}{2(xy+xz+yz)}-2=\frac{4x^2+(y-z)^2}{2(xy+xz+yz)}\geq0$ .

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kk108

#39 h Aug 3, 2016, 5:34 AM • 2 Y

Y by Adventure10, Mango247

anantmudgal09 wrote:

In turn, we could bound the variables in $[0,4]^3$ and see that by Lagrange Multipliers we are done.

What are Langrange Multipliers ?

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arqady

#40 h Aug 3, 2016, 6:26 AM • 2 Y

Y by Adventure10, Mango247

kk108 wrote:

What are Langrange Multipliers ?

See here:
https://en.wikipedia.org/wiki/Lagrange_multiplier

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sqing

#41 h Aug 3, 2016, 6:46 AM • 1 Y

Y by Adventure10

arqady wrote:

sqing wrote:

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $4a+b+c\ge 2.$

Let $a=\frac{x}{y+z}$ and $b=\frac{y}{x+z}$ , where $x$ , $y$ and $z$ are positives.
Hence, $c=\frac{z}{x+y}$ and by C-S we obtain:
$4a+b+c-2=\frac{4x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}-2\geq\frac{(2x+y+z)^2}{2(xy+xz+yz)}-2=\frac{4x^2+(y-z)^2}{2(xy+xz+yz)}\geq0$ .

Very nice.

Proof of matha:
The inequality is strict unless the pronunciation saying "non-negative".
There are positive $\displaystyle {a, b, c}$ that
$\displaystyle {x = \frac {a} {b + c}, y = \frac {b} {c + a}, z = \frac {c} {a + b},}$ when the verifiable written
$\displaystyle {\frac {4a} {b + c} + \frac {b} {c + a} + \frac {c} {a + b} \geq 2.}$
From Cauchy-Schwarz is
$\displaystyle{\frac{4a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{4a^2}{ab+ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ac+bc}\geq \frac {(2a + b + c) ^ 2} {2 (ab + bc + ca)}}$
it is sufficient to prove that
$\displaystyle {(2a + b + c) ^ 2 \geq 4 (ab + bc + ca)}$
which is equivalent to the apparent $\displaystyle {4a ^ 2 + (b-c) ^ 2 \geq 0.}$

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sqing

#42 h Aug 3, 2016, 10:28 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $4a+b+c\ge 2.$ (Mathematical Olympiad 2015 A.Petroupolis Hg (9th grade))

Attachments:

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Grotex

#43 h Aug 3, 2016, 3:33 PM • 1 Y

Y by Adventure10

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $a^2+b+c\ge \frac{5}{4}.$ (By Grotex)

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sqing

#44 h Aug 3, 2016, 10:17 PM • 2 Y

Y by Adventure10, Mango247

Grotex wrote:

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $a^2+b+c\ge \frac{5}{4}.$ (By Grotex)

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#45 h Aug 4, 2016, 3:11 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\frac 1{8a^2+1}+\frac 1{8b^2+1}+\frac 1{8c^2+1}\ge 1.$

Let $t = \sqrt[3]{{abc}}$ ，Since $1 = ab + bc + ca + 2abc \ge 3\sqrt[3]{{\left( {abc} \right)^2 }} + 2abc = 3t^2 + 2t^3 \Rightarrow 2t^3 + 3t^2 - 1 \le 0 \Rightarrow$
$\left( {2t - 1} \right)\left( {t + 1} \right)^2 \le 0 \Leftrightarrow t \le \frac{1}{2} \Leftrightarrow abc \le \frac{1}{8}$ ，
Since $abc \le \frac{1}{8} \Rightarrow \exists k,x,y,z > 0:\left( {a^2 ,b^2 ,c^2 } \right) = \left( {\frac{{kyz}}{{x^2 }},\frac{{kzx}}{{y^2 }},,\frac{{kxy}}{{z^2 }}} \right)$ ,and $0 < k \le \frac{1}{4}$ ，
By C-S We have
$\sum {\frac{1}{{8a^2 + 1}}} = \sum {\frac{{x^2 }}{{x^2 + 8kyz}}} = \ge \frac{{\left( {\sum x } \right)^2 }}{{\sum {x^2 + 24kxyz} }} \ge \frac{{\sum {x^2 + 6xyz} }}{{\sum {x^2 + 24 \times \frac{1}{4}xyz} }} = 1$

Generalization
Let $a,b,c,\lambda ,\mu$ are positive real numbers such that $ab + bc + ca + 2abc = 1$ and $\lambda \le 2^{\mu + 1}$ .Prove that
$\frac{1}{{\lambda a^\mu + 1}} + \frac{1}{{\lambda b^\mu + 1}} + \frac{1}{{\lambda c^\mu + 1}} \ge 1{\rm{ }}$

This post has been edited 2 times. Last edited by scpajmb, Aug 4, 2016, 3:24 AM

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sqing

#46 h Feb 14, 2019, 1:20 PM • 1 Y

Y by Adventure10

sqing wrote:

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ Prove that $4a+b+c\ge 2.$ (Mathematical Olympiad 2015 A.Petroupolis Hg (9th grade))

Let $a,b,c$ be positiv real numbers such that $ab+bc+ca+2abc=1.$ For any positive numbers $x,y,z,$ prove that $xa+yb+zc\ge \frac 12(\sqrt{x}+\sqrt{y}+\sqrt{z})^2-(x+y+z).$

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sqing

#47 h Feb 28, 2019, 2:47 AM • 1 Y

Y by Adventure10

huricane wrote:

Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.

Let $x,y,z$ be three positive real numbers such that $x^2+y^2+z^2+3=2(xy+yz+zx) .$ Show that
$\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\ge 3,$ Stars of Mathematics 2017, Juniors

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sqing

#48 h Apr 21, 2023, 9:06 AM

Y by

huricane wrote:

Let $a,b,c\ge 0$ be three real numbers such that $ab+bc+ca+2abc=1.$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.

Let $a,b,c \ge 0 : ab+bc+ca=1$ . Prove that:
$\sqrt{a}+\sqrt{b}+\sqrt{c} \ge 2$

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