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#1 h May 4, 2015, 1:29 PM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c \geq 0$ . Prove that:
$\frac{a(b+c)}{a^2+2bc}+\frac{b(c+a)}{b^2+2ac}+\frac{c(a+b)}{c^2+2ab} \geq 1+\frac{ab+ac+bc}{a^2+b^2+c^2}$

Hoa nhài

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283 posts

Wangzu

#2 h May 5, 2015, 8:16 AM • 3 Y

Y by A_Gappus, Adventure10, Mango247

We have the ineq for all non-negative:

$\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\geq \frac{2}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}$
$\Leftrightarrow (ab+bc+ac)(\sum \frac{1}{a^2+2bc})\geq 2+\frac{ab+bc+ac}{a^2+b^2+c^2}$
$\Leftrightarrow \sum \frac{a(b+c)}{a^2+2bc}+\frac{bc}{a^2+2bc}\geq 2+\frac{ab+bc+ac}{a^2+b^2+c^2}$

And then we have to prove the ineq:
$\sum \frac{bc}{a^2+2bc}\leq 1\Leftrightarrow \sum \frac{a^2}{a^2+2bc}\geq 1$ which just by C-S

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Wangzu

#3 h May 5, 2015, 8:24 AM • 2 Y

Y by Adventure10, Mango247

I have another idea by S.O.S, our inequality equivalent to:

$S_{a}(b-c)^2+S_{b}(a-c)^2+S_{c}(a-b)^2\geq 0$

And $S_{a}=\frac{3}{2(a^2+b^2+c^2)}-\frac{4ab+4ac-4a^2-bc}{(b^2+2ac)(c^2+2ab)}$
$S_{b}=\frac{3}{2(a^2+b^2+c^2)}-\frac{4bc+4ab-4b^2-ac}{(a^2+2bc)(c^2+2ab)}$
$S_{c}=\frac{3}{2(a^2+b^2+c^2)}-\frac{4ac+4bc-4c^2-ab}{(a^2+2bc)(b^2+2ac)}$

Can someone prove this(by this idea)?

This post has been edited 1 time. Last edited by Wangzu, May 5, 2015, 8:29 AM

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xzlbq

#4 h May 5, 2015, 9:56 AM • 4 Y

Y by Wangzu, dragonheart6, Adventure10, Mango247

BuiBaAnh wrote:

Let $a,b,c \geq 0$ . Prove that:
$F=\frac{a(b+c)}{a^2+2bc}+\frac{b(c+a)}{b^2+2ac}+\frac{c(a+b)}{c^2+2ab}-1-\frac{ab+ac+bc}{a^2+b^2+c^2}\geq 0$

Hoa nhài

a SOS is:

$f={\frac {Q}{ \left( {c}^{2}+2\,ab \right) \left( {b}^{2}+2\,ac \right) \left( {a}^{2}+2\,bc \right) \left( {a}^{2}+{b}^{2}+{c}^{2} \right) }}$

$Q= \left( 2\,a-b-c \right) ^{2}a{b}^{2}{c}^{3}+ \left( 2\,b-a-c \right) ^{2}b{c}^{2}{a}^{3}+ \left( 2\,c-a-b \right) ^{2}c{a}^{2}{b}^ {3}+1/2\, \left( -a{b}^{2}+{c}^{2}a-abc+{b}^{2}c \right) ^{2}{a}^{2}+1 /2\, \left( {a}^{2}b-b{c}^{2}-abc+{c}^{2}a \right) ^{2}{b}^{2}+1/2\, \left( {a}^{2}b-abc-{a}^{2}c+{b}^{2}c \right) ^{2}{c}^{2}+ \left( c-b \right) ^{2} \left( a-b \right) ^{2}{c}^{2}ab+ \left( b-a \right) ^{2 } \left( c-a \right) ^{2}{b}^{2}ca+ \left( a-c \right) ^{2} \left( b-c \right) ^{2}{a}^{2}bc+1/2\, \left( a-b \right) ^{2} \left( a+b-2\,c \right) ^{2}{c}^{4}+1/2\, \left( b-c \right) ^{2} \left( b-2\,a+c \right) ^{2}{a}^{4}+1/2\, \left( c-a \right) ^{2} \left( a-2\,b+c \right) ^{2}{b}^{4}$

BQ

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283 posts

Wangzu

#5 h May 5, 2015, 10:36 AM • 2 Y

Y by Adventure10, Mango247

I have found this following inequality(may be easy):

For all positive real numbers,we have:

$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 1+\frac{16abc}{(a+b)(b+c)(a+c)}$

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2032 posts

#6 h May 5, 2015, 11:48 AM • 2 Y

Y by Adventure10, Mango247

For #5
We have
$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}- [1+\frac{16abc}{(a+b)(b+c)(a+c)}]=$
$\sum{\frac{(a-c)^2(b-c)^2}{(c^2+ab)(b+c)(c+a)}}\ge{0}$

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lumia

#7 h May 5, 2015, 2:59 PM • 1 Y

Y by Adventure10

Wangzu wrote:

I have found this following inequality(may be easy):

For all positive real numbers,we have:

$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 1+\frac{16abc}{(a+b)(b+c)(a+c)}$

hello, I know it has a solution by using Am-Gm in a book of VQBCan and TQAnh

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30149 posts

arqady

#8 h May 23, 2015, 9:40 AM • 3 Y

Y by luofangxiang, Adventure10, Mango247

Let $a$ , $b$ and $c$ be positive numbers. Prove that:
$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 2+\frac{3abc(a+b+c)}{(a^2+b^2+c^2)^2}$

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30149 posts

arqady

#9 h May 23, 2015, 10:03 AM • 2 Y

Y by Adventure10, Mango247

Wangzu wrote:

For all positive real numbers,we have:
$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 1+\frac{16abc}{(a+b)(b+c)(a+c)}$

$\sum_{cyc}\frac{a(b+c)}{a^2+bc}-1-\frac{16abc}{(a+b)(b+c)(a+c)}=2-\frac{16abc}{(a+b)(b+c)(a+c)}-\sum_{cyc}\left(1-\frac{a(b+c)}{a^2+bc}\right)=$
$=\frac{2\sum\limits_{cyc}(a^2b+a^2c-2abc)}{(a+b)(a+c)(b+c)}-\sum_{cyc}\frac{(a-b)(a-c)}{a^2+bc}=$
$=\frac{2\sum\limits_{cyc}(b+c)(a-b)(a-c)}{(a+b)(a+c)(b+c)}-\sum_{cyc}\frac{(a-b)(a-c)}{a^2+bc}=\sum_{cyc}\frac{(a-b)^2(a-c)^2}{(a+b)(a+c)(a^2+bc)}\geq0$ .

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2032 posts

#10 h May 24, 2015, 2:32 AM • 1 Y

Y by Adventure10

For #8
We have
$[\sum{\frac{a(b+c)}{a^2+bc}}-2][\sum{a^2}]^2-3abc\sum{a}=\sum{\frac{ab(2a^2+ab+2b^2)(ab+c^2)(a-b)^2}{2(a^2+bc)(ac+b^2)}}$
$+\frac{(a-b)^2(b-c)^2(c-a)^2}{2(a^2+bc)(ac+b^2)(ab+c^2)}\sum{(a^2-ab+b^2)^2}\ge{0}$

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4613 posts

#11 h Nov 30, 2016, 10:08 AM • 2 Y

Y by teomihai, Adventure10

arqady wrote:

Let $a$ , $b$ and $c$ be positive numbers. Prove that:
$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 2+\frac{3abc(a+b+c)}{(a^2+b^2+c^2)^2}$

//cdn.artofproblemsolving.com/images/9/7/0/97089c0e44943dbe33c6f1d7a85c6aceaaa41e66.jpg

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1119 posts

#12 h Jun 2, 2020, 2:19 PM • 2 Y

Y by teomihai, Mango247

arqady wrote:

Let $a$ , $b$ and $c$ be positive numbers. Prove that:
$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\geq 2+\frac{3abc(a+b+c)}{(a^2+b^2+c^2)^2}$

$LHS-RHS=\frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2+bc)(b^2+ca)(c^2+ab)}+\frac{abc[\sum ab(a+b)-abc]\sum (a^2-b^2)^2}{2(a^2+bc)(b^2+ca)(c^2+ab)(a^2+b^2+c^2)^2}\ge 0$

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