An inequality with condition sum a/(1+a)=2

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An inequality with condition sum a/(1+a)=2

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Cauchy Inequality

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Source: Middle European Mathematical Olympiad 2011 - Team Compt. T-2

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#1 h Sep 6, 2011, 11:37 AM • 2 Y

Y by Adventure10, Mango247

Let $a, b, c$ be positive real numbers such that
$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2.$
Prove that
$\frac{\sqrt a + \sqrt b+\sqrt c}{2} \geq \frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}.$

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#2 h Sep 7, 2011, 3:21 AM • 2 Y

Y by LoveMath4ever, Adventure10

Maybe this can help, but I can't finish..

The condition $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2$ is equivalent to $abc= a+b+c+2$ and therefore we can make the substitution

$a=\frac{y+z}{x}, b=\frac{x+z}{y}$ and $c=\frac{x+y}{z}$

Hence the inequality becomes ${ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right)$

Any thoughts on this?

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anonymouslonely

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#3 h Sep 7, 2011, 4:26 PM • 2 Y

Y by Adventure10, Mango247

I think now you can use Cebisev using $y+z-x$ is decreasing in $x$ and $x(y+z)$ is increasing in $x$ .

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#4 h Sep 7, 2011, 5:22 PM • 7 Y

Y by mudok, chinacai, buganalupin, lazizbek42, Adventure10, and 2 other users

hatchguy wrote:

${ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right)$

$\Leftrightarrow \sum \left ( \sqrt{\frac{y+z}{2x}} - \sqrt{\frac{2x}{y+z}} \right ) \geq 0$

$\Leftrightarrow \sum \left ( \frac{y+z-2x}{\sqrt{x(y+z)}} \right ) \geq 0 \Leftrightarrow \sum \left ( \frac{y-x}{\sqrt{x(y+z)}} + \frac{x-y}{\sqrt{y(x+z)}} \right ) \geq 0$

$= \sum \left ( (x-y)(\frac{1}{\sqrt{y(z+x)}} - \frac{1}{\sqrt{x(y+z)}}) \right )$

$= \sum \left ( \frac{z(x-y)^2}{ \left [\sqrt{y(z+x)}+\sqrt{x(y+z)} \right ] \sqrt{xy(z+x)(z+y)} } \right ) \geq 0$

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#5 h Sep 7, 2011, 6:44 PM • 2 Y

Y by Adventure10, Mango247

Nice (end of )solution Mahanmath!.
PS

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#6 h Sep 7, 2011, 8:23 PM • 2 Y

Y by Adventure10, Mango247

amparvardi wrote:

Let $a, b, c$ be positive real numbers such that
$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2.$
Prove that
$\frac{\sqrt a + \sqrt b+\sqrt c}{2} \geq \frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}.$

This is solution that I found during the competition:

Proof

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#7 h Sep 7, 2011, 9:46 PM • 2 Y

Y by Adventure10, Mango247

hatchguy wrote:

Hence the inequality becomes ${ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right)$

Any thoughts on this?

Substitute $r=\sqrt{x}, s=\sqrt{y}, t=\sqrt{z}$ and use $\sqrt{r^2+s^2}\ge \frac1 {\sqrt{2}}(r+s)$ in the numerators as well as the denominators to get:

$\sum \frac{r+s}t \ge 4 \sum \frac r {s+t}$

This inequality has been discussed before and is very easy

.

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#8 h Sep 9, 2011, 1:02 PM • 2 Y

Y by Adventure10, Mango247

why are you hiting your head on the desk ?
$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2.$
$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1.$
so $a=\frac{y+z}{x}, b=\frac{x+z}{y}$ and $c=\frac{x+y}{z}$ ,

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#9 h Sep 11, 2011, 12:40 PM • 2 Y

Y by Adventure10, Mango247

hatchguy wrote:

${ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right)$

Generalisation:
If $x_1,x_2,\cdots,x_n>0,S=x_1+x_2+\cdots+x_n,n\ge 2$ ,then
$\sqrt{\frac{S-x_1}{x_1}}+\sqrt{\frac{S-x_2}{x_2}}+\cdots+\sqrt{\frac{S-x_n}{x_n}} \ge$
$\ge (n-1) \left(\sqrt{\frac{x_1}{S-x_1}}+\sqrt{\frac{x_2}{S-x_2}}+\cdots+\sqrt{\frac{x_n}{S-x_n}}\right)$

Proof,notice that
(1) ${\frac{S-x_1}{x_1}+\frac{S-x_2}{x_2}+\cdots+\frac{S-x_n}{x_n}}\ge$
$\ge (n-1)^2 \left(\frac{x_1}{S-x_1}+\frac{x_2}{S-x_2}+\cdots+\frac{x_n}{S-x_n}\right)$

This is because,By cauchy inequality,
$(n-1)^2\frac{x_1}{S-x_1}\le \frac{x_1}{x_2}+\frac{x_1}{x_3}+\cdots+\frac{x_1}{x_n}$
And the other n-1 similar inequalies,then sum them up.

(2)For $1\le i<j\le n$ ,we have:
$\sqrt{\frac{(S-x_i)(S-x_j)}{x_ix_j}}\ge 1+\sum_{k=1,k \neq i,j}\frac{x_k}{\sqrt{x_ix_j}}$

(3)For $1\le i<j\le n$ ,we have:
$\sqrt{\frac{x_ix_j}{(S-x_i)(S-x_j)}}\le \frac{1}{2}\left(\frac{x_i}{S-x_j}+\frac{x_j}{S-x_i}\right)$

By,(1),(2),(3).We can square the orginal inequality,then not hard to conclude a proof.

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#10 h Sep 18, 2011, 6:33 AM • 2 Y

Y by Adventure10, Mango247

2011年中欧数学奥林匹克不等式试题的一般结论

Attachments:

2011年中欧数学奥林匹克不等式试题的一般结论.pdf (60kb)

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#11 h Jul 23, 2014, 12:23 AM • 2 Y

Y by Adventure10, Mango247

Georg-A. wrote:

hatchguy wrote:

Hence the inequality becomes ${ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right)$

Any thoughts on this?

Substitute $r=\sqrt{x}, s=\sqrt{y}, t=\sqrt{z}$ and use $\sqrt{r^2+s^2}\ge \frac1 {\sqrt{2}}(r+s)$ in the numerators as well as the denominators to get:

$\sum \frac{r+s}t \ge 4 \sum \frac r {s+t}$

This inequality has been discussed before and is very easy

.

The inequality $\sum \frac{r+s}t \ge 4 \sum \frac r {s+t}$ not equivalent ${ \sqrt \frac{y+z}{x}+\sqrt \frac{x+z}{y}+\sqrt \frac{x+y}{z}}\geq2\left({\sqrt \frac{x}{y+z}}+{\sqrt \frac{y}{x+z}}+{\sqrt \frac{z}{x+y}}\right)$

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sqing

#12 h Jul 23, 2014, 5:18 AM • 1 Y

Y by Adventure10

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=114791&hilit=inequality
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2869656

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#14 h Jul 9, 2019, 12:39 AM • 2 Y

Y by Adventure10, Mango247

Amir Hossein wrote:

Let $a, b, c$ be positive real numbers such that $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2.$ Prove that
$\frac{\sqrt a + \sqrt b+\sqrt c}{2} \geq \frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}.$

Let $a, b, c$ be positive real numbers such that $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2.$ Prove that
$a^2 +b^2+c^2 \geq 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}).$ $abc+4\geq 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}).$

This post has been edited 2 times. Last edited by sqing, Jul 9, 2019, 12:43 AM

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#15 h Oct 4, 2019, 2:39 AM • 1 Y

Y by Adventure10

For three positive real numbers a, b, c, prove that
$\sqrt{\frac{a+b}{c}} +\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}{ \geq } 2(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}})$ https://artofproblemsolving.com/community/c6h397101p2208537
https://artofproblemsolving.com/community/c6h475161p2661111
https://artofproblemsolving.com/community/c6h511112p2869656

Amir Hossein wrote:

Let $a, b, c$ be positive real numbers such that
$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2.$ Prove that
$\frac{\sqrt a + \sqrt b+\sqrt c}{2} \geq \frac{1}{\sqrt a}+\frac{1}{\sqrt b}+\frac{1}{\sqrt c}.$

$cosA+cosB+cosC\geq 2(cosBcosC+cosCcosA+cosAcosB)$

Attachments:

This post has been edited 2 times. Last edited by sqing, Jun 4, 2020, 3:06 AM

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