Simple inequality (86)

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Simple inequality (86)

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sqing

#1 h Apr 23, 2013, 5:51 AM • 2 Y

Y by Adventure10, Mango247

Let positive real numbers $a,b,c$ . Show that $\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} >1 .$

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arqady

#2 h Apr 23, 2013, 11:37 AM • 3 Y

Y by sqing, Adventure10, Mango247

sqing wrote:

Let positive real numbers $a,b,c$ . Show that $\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} >1 .$

$\left(\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}}\right)^2\sum_{cyc}b^2c^2(a+b)(a+c)\geq(ab+ac+bc)^3$ .
Thus, it remains to prove that $(ab+ac+bc)^3>\sum_{cyc}b^2c^2(a^2+ab+ac+bc)$ , which is obvious.
The following inequality is also true and also simple.
For non-negatives $a$ , $b$ and $c$ such that $ab+ac+bc\neq0$ prove that:
$\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}} \geq\sqrt{1+\frac{10abc}{(a+b)(a+c)(b+c)}}$

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sqing

#3 h Apr 24, 2013, 7:03 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c\geq 0$ , prove that
$\sqrt{2}+\frac{4(3-2\sqrt{2})abc}{(a+b)(b+c)(c+a)}\geq \sum_{cyc}{\sqrt{\frac{ab}{(a+c)(b+c)}}}\geq 1+\dfrac{4abc}{(a+b)(b+c)(c+a)}$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2836557#p2836557

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Sayan

#4 h Apr 24, 2013, 11:32 AM • 2 Y

Y by Adventure10, Mango247

Arqady's version is stronger,
we have:
For non-negatives $a$ , $b$ and $c$ such that $ab+ac+bc\neq0$ prove that:
$\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}} \geq\sqrt{1+\frac{10abc}{(a+b)(a+c)(b+c)}}\geq 1+\dfrac{4abc}{(a+b)(b+c)(c+a)}$

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sqing

#5 h Apr 24, 2013, 11:14 PM • 3 Y

Y by lminsl, Adventure10, Mango247

Thank Sayan.
For non-negatives $a$ , $b$ and $c$ such that $ab+ac+bc\neq0$ prove that: $\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} \geq 1$

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#6 h Apr 24, 2013, 11:47 PM • 2 Y

Y by Adventure10, Mango247

a solution for the inequality

Attachments:

Using inequality xy.zip (8kb)

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gxggs

#7 h Apr 25, 2013, 3:31 AM • 2 Y

Y by Adventure10, Mango247

Sayan wrote:

Arqady's version is stronger,
we have:
For non-negatives $a$ , $b$ and $c$ such that $ab+ac+bc\neq0$ prove that:
$\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}} \geq\sqrt{1+\frac{10abc}{(a+b)(a+c)(b+c)}}\geq 1+\dfrac{4abc}{(a+b)(b+c)(c+a)}$

Here is the sharper one:
$\sum_{cyc}{\sqrt{\frac{bc}{(a+b)(a+c)}}}\geq \sqrt{1+\sqrt{\frac{25abc}{2(a+b)(b+c)(c+a)}}}$

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arqady

#8 h Apr 25, 2013, 3:58 AM • 2 Y

Y by sqing, Adventure10

For the proof of the last inequality we can use $\sum_{cyc}\sqrt{a(a+b)(a+c)}\geq2\sqrt{(a+b+c)(ab+ac+bc)}$ .

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sqing

#9 h Sep 14, 2013, 1:29 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be positive real numbers . Show that $\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} \le \frac{3}{2} .$

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Sayan

#10 h Sep 14, 2013, 2:37 AM • 1 Y

Y by Adventure10

sqing wrote:

Let $a,b,c$ be positive real numbers . Show that $\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} \le \frac{3}{2} .$

We can use just AM-GM:
$\begin{align*}\sum\sqrt{\frac{bc}{(c+a)(a+b)}} & \le \frac12\sum\left(\frac{c}{c+a}+\frac{b}{a+b}\right) \\ & =\frac12\sum\left(\frac{a}{a+b}+\frac{b}{a+b}\right)=\frac32\end{align*}$

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sqing

#11 h Oct 14, 2015, 12:48 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be positive real numbers . Show that $\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \geq1 .$

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arqady

#12 h Oct 14, 2015, 4:12 AM • 1 Y

Y by Adventure10

sqing wrote:

Let $a,b,c$ be positive real numbers . Show that $\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \geq1$

By Holder $\left(\sum_{cyc}\sqrt{\frac{ab}{(b+2c)(2c+a)}}\right)^2\sum_{cyc}a^2b^2(a+2c)(b+2c)\geq(ab+ac+bc)^3$ .
Hence, it remains to prove that $(ab+ac+bc)^3\geq\sum_{cyc}a^2b^2(a+2c)(b+2c)$ , which is $abc\sum_{cyc}c(a-b)^2\geq0$ . Done!

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Wangzu

#13 h Oct 14, 2015, 8:32 AM • 2 Y

Y by Adventure10, Mango247

For the folowing inequality: $\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}} \geq\sqrt{1+\frac{10abc}{(a+b)(a+c)(b+c)}}$
By squaring the both sides we need to show that:

$\sum \frac{ab}{(b+c)(c+a)}+2\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\left(\sum \sqrt{\frac{a}{b+c}}\right)\geq 1+\frac{10abc}{(a+b)(b+c)(c+a)}$ If we use the well known inequality:

$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq 2\sqrt{1+\frac{abc}{(a+b)(b+c)(c+a)}}$
Then we need to prove that:

$(a+b)(b+c)(c+a) \geq 8abc$ which is obvius by AM-GM

This post has been edited 4 times. Last edited by Wangzu, Oct 14, 2015, 4:39 PM
Reason: ok

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#14 h Oct 14, 2015, 12:05 PM • 2 Y

Y by Adventure10, Mango247

Dear Wangzu
Sorry, can you write fuller solution here?

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Wangzu

#15 h Oct 14, 2015, 4:46 PM • 2 Y

Y by Adventure10, Mango247

gxggs wrote:

Sayan wrote:

Arqady's version is stronger,
we have:
For non-negatives $a$ , $b$ and $c$ such that $ab+ac+bc\neq0$ prove that:
$\sum_{cyc}\sqrt{\frac{bc}{(c+a)(a+b)}} \geq\sqrt{1+\frac{10abc}{(a+b)(a+c)(b+c)}}\geq 1+\dfrac{4abc}{(a+b)(b+c)(c+a)}$

Here is the sharper one:
$\sum_{cyc}{\sqrt{\frac{bc}{(a+b)(a+c)}}}\geq \sqrt{1+\sqrt{\frac{25abc}{2(a+b)(b+c)(c+a)}}}$

By squaring the both sides, we need prove:

$\sum \frac{ab}{(b+c)(c+a)}+2\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\left(\sum \sqrt{\frac{a}{b+c}}\right)\geq 1+5\sqrt{\frac{abc}{2(a+b)(b+c)(c+a)}}$
We still can use the inequality:

$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq 2\sqrt{1+\frac{abc}{ (a+b)(b+c)(c+a)}}$
Then let $x=\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}, x\leq \frac{1}{2\sqrt{2}}$ we got something obvious.

Or we can use the folowing:

$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}+2(2\sqrt{2}-3)\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\geq 2$

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sqing

#16 h Oct 16, 2015, 1:28 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be positive real numbers . Show that $\frac{a^3}{(c+2a)(2a+b)}+\frac{b^3}{(a+2b)(2b+c)}+\frac{c^3}{(b+2c)(2c+a)}\geq \frac{a+b+c}{9}.$

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arqady

#17 h Oct 16, 2015, 1:57 AM • 2 Y

Y by Adventure10, Mango247

Let $a$ , $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$ . Prove that:
$\frac{1}{3}\sqrt{\frac{a^2+b^2+c^2}{3}}\leq\frac{a^3}{(c+2a)(2a+b)}+\frac{b^3}{(a+2b)(2b+c)}+\frac{c^3}{(b+2c)(2c+a)}\leq \frac{a^2+b^2+c^2}{3(a+b+c)}$

This post has been edited 1 time. Last edited by arqady, Oct 16, 2015, 2:02 AM

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sqing

#18 h Oct 16, 2015, 2:07 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be non-negative real numbers such that $ab+ac+bc\neq0$ . Show that $\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}.$

This post has been edited 1 time. Last edited by sqing, Oct 16, 2015, 2:09 AM

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Wangzu

#19 h Oct 16, 2015, 2:51 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a,b,c$ be positive real numbers . Show that $\frac{a^3}{(c+2a)(2a+b)}+\frac{b^3}{(a+2b)(2b+c)}+\frac{c^3}{(b+2c)(2c+a)}\geq \frac{a+b+c}{9}.$

It's just AM-GM for: $\frac{a^3}{(c+2a)(2a+b)}+\frac{c+2a}{27}+\frac{2a+b}{27} \geq \frac{a}{3}$

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sqing

#20 h Oct 16, 2015, 2:58 AM • 2 Y

Y by Adventure10, Mango247

Wangzu wrote:

sqing wrote:

Let $a,b,c$ be positive real numbers . Show that $\frac{a^3}{(c+2a)(2a+b)}+\frac{b^3}{(a+2b)(2b+c)}+\frac{c^3}{(b+2c)(2c+a)}\geq \frac{a+b+c}{9}.$

It's just AM-GM for: $\frac{a^3}{(c+2a)(2a+b)}+\frac{c+2a}{27}+\frac{2a+b}{27} \geq \frac{a}{3}$

Yeah .
Thanks.

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arqady

#21 h Oct 16, 2015, 7:03 AM • 3 Y

Y by Monreal, Adventure10, Mango247

sqing wrote:

Let $a,b,c$ be non-negative real numbers such that $ab+ac+bc\neq0$ . Show that $\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}.$

By C-S $\left(\sum_{cyc}\sqrt{\frac{bc}{(c+2a)(2a+b)}}\right)^2\leq\sum_{cyc}\frac{c}{c+2a}\sum_{cyc}\frac{b}{b+2a}\leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}\cdot \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}=\frac{a^2+b^2+c^2}{ab+ac+bc}$ .

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math90

#22 h Oct 16, 2015, 7:26 AM • 2 Y

Y by arqady, Adventure10

sqing wrote:

Let $a,b,c$ be positive real numbers . Show that $\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \geq1 .$

$\sum_{cyc}\sqrt{\frac{bc}{(2a+b)(2a+c)}}=\sum_{cyc}\frac{bc}{\sqrt{b(2a+c)}\sqrt{c(2a+b)}}\geq\sum_{cyc}\frac{2bc}{b(2a+c)+c(2a+b)}=1.$

This post has been edited 2 times. Last edited by math90, Oct 16, 2015, 7:35 AM

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sqing

#23 h Oct 16, 2015, 7:40 AM • 1 Y

Y by Adventure10

Very very nice .
Thank you very much .

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#24 h Oct 16, 2015, 9:29 AM • 1 Y

Y by Adventure10

arqady wrote:

sqing wrote:

Let $a,b,c$ be non-negative real numbers such that $ab+ac+bc\neq0$ . Show that $\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}.$

By C-S $\left(\sum_{cyc}\sqrt{\frac{bc}{(c+2a)(2a+b)}}\right)^2\leq\sum_{cyc}\frac{c}{c+2a}\sum_{cyc}\frac{b}{b+2a}\leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}\cdot \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}=\frac{a^2+b^2+c^2}{ab+ac+bc}$ .

Do you have a nice proof for $\sum_{cyc}\frac{c}{c+2a} \leq \sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}}$ . Thanks!

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arqady

#25 h Oct 16, 2015, 2:20 PM • 1 Y

Y by Adventure10

I have no a nice proof. $uvw$ kills it.

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#26 h Mar 28, 2016, 12:20 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

Let $a$ , $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$ . Prove that:
$\frac{1}{3}\sqrt{\frac{a^2+b^2+c^2}{3}}\leq\frac{a^3}{(c+2a)(2a+b)}+\frac{b^3}{(a+2b)(2b+c)}+\frac{c^3}{(b+2c)(2c+a)}\leq \frac{a^2+b^2+c^2}{3(a+b+c)}$

//cdn.artofproblemsolving.com/images/f/d/8/fd8f0c18b4662eb28a2896c190ec51386ceeb3db.jpg

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#27 h Mar 28, 2016, 1:08 PM • 2 Y

Y by Adventure10, Mango247

$\sqrt{\sum_{cyc}a^2\sum_{cyc}(a^2+bc)^2}\leq 6$
$\iff \sum_{cyc}(a^2+bc)^2\leq 12$
Can you elaborate the above inequality ?

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#28 h Mar 28, 2016, 2:04 PM • 2 Y

Y by Adventure10, Mango247

//cdn.artofproblemsolving.com/images/2/6/2/26244ef31719e91a1d1e827074d8826e0473dc8a.jpg

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#29 h Apr 12, 2016, 7:38 AM • 1 Y

Y by Adventure10

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#31 h Apr 12, 2016, 8:44 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let positive real numbers $a,b,c$ . Show that $\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} >1 .$

Let $x=\frac{a}{b+c}$ , $y=\frac{b}{a+c}$ , $z=\frac{c}{b+a}$ . Then $xy+yz+xz+2xyz=1$ and we need to prove $\sqrt{xy}+\sqrt{yz}+\sqrt{zx}>1$ . Now, let
$m=xy$ , $n=yz$ , $p=zx$ . Then we need to prove $\sqrt{m}+\sqrt{n}+\sqrt{p}>1$ with $m+n+p+2\sqrt{mnp}=1$ . We have that $\sqrt{m}+\sqrt{n}+\sqrt{p}>1$ is equivalent with $m+n+p+2(\sqrt{mn}+\sqrt{np}+\sqrt{pm})>1=m+n+p+2\sqrt{mnp}$ . So we want to prove:
$\sqrt{mn}+\sqrt{np}+\sqrt{pm}>\sqrt{mnp}$ . Note that $\sqrt{mn}+\sqrt{np}+\sqrt{pm} \geq 3*(mnp)^{\frac{1}{3}}$ so we want to prove $3*(mnp)^{\frac{1}{3}}>\sqrt{mnp}$ or $3^6*(mnp)^2>(mnp)^3$ or $mnp<3^6$ which is true because $m+n+p+2\sqrt{mnp}=1 \geq 3*(mnp)^{\frac{1}{3}}+2\sqrt{mnp}$ . Q.E.D

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sqing

#32 h Aug 28, 2020, 7:54 AM

Y by

sqing wrote:

Let $a,b,c$ be positive real numbers . Show that $\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} \le \frac{3}{2} .$

Let positive real numbers $a,b,c$ . Show that
$\sqrt{\frac{bc}{(c+a)(a+b)}}+\sqrt{\frac{ca}{(a+b)(b+c)}}+\sqrt{\frac{ab}{(b+c)(c+a)}} \leq\frac{3}{2}-\frac{(b-c)^2}{12(b+c)^2} .$

sqing wrote:

Let $a,b,c$ be positive real numbers . Show that $\sqrt{\frac{bc}{(c+2a)(2a+b)}}+\sqrt{\frac{ca}{(a+2b)(2b+c)}}+\sqrt{\frac{ab}{(b+2c)(2c+a)}} \geq1 .$

N:
$\sqrt {{\frac {bc}{ \left(c+2a\right) \left( 2a+b\right) }}}\geq { \frac {bc}{ab+bc+ca}}$

This post has been edited 1 time. Last edited by sqing, Aug 28, 2020, 7:57 AM

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