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#1 h Nov 11, 2009, 2:24 PM • 2 Y

Y by Adventure10, Mango247

For all $a,b,c$ be nonnegative real numbers, show that
$\frac{3(ab+bc+ca)}{2(a^2b^2+b^2c^2+c^2a^2)}\le\frac1{a^2+bc}+\frac1{b^2+ca}+\frac1{c^2+ab}\le\frac{3(a^2+b^2+c^2)}{2(a^2b^2+b^2c^2+c^2a^2)}$

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#2 h Nov 11, 2009, 3:11 PM • 2 Y

Y by Adventure10, Mango247

dduclam wrote:

For all $a,b,c$ be nonnegative real numbers, show that
$\frac {3(ab + bc + ca)}{2(a^2b^2 + b^2c^2 + c^2a^2)}\le\frac1{a^2 + bc} + \frac1{b^2 + ca} + \frac1{c^2 + ab}\le\frac {3(a^2 + b^2 + c^2)}{2(a^2b^2 + b^2c^2 + c^2a^2)}$

assume $a = max{(a,b,c)}$

$\frac1{a^2 + bc} + \frac1{b^2 + ca} + \frac1{c^2 + ab}\le\frac {3(a^2 + b^2 + c^2)}{2(a^2b^2 + b^2c^2 + c^2a^2)}$

is equivalent to

$(3a^4bc + (b + c)(b^2 + c^2)a^3 + ( - b^4 - c^4 + 3b^3c + b^2c^2 + 3bc^3)a^2 + 2bc(b + c)(b^2 - bc + c^2)a + bc(b^2 + 3bc + c^2)(b - c)^2)(a - b)(a - c)$
$+ (4b^4ca + 6b^3c^2a + 6b^2c^3a + 4bc^4a - b^4c^2 - 2b^3c^3 - c^4b^2)(b - c)^2$

$3a^4bc + (b + c)(b^2 + c^2)a^3 + ( - b^4 - c^4 + 3b^3c + b^2c^2 + 3bc^3)a^2 + 2bc(b + c)(b^2 - bc + c^2)a + bc(b^2 + 3bc + c^2)(b - c)^2\geq (b + c)(b^2 + c^2)a^3 + ( - b^4 - c^4 )a^2\geq a^2((b^3 + c^3)a - (b^4 + c^4))\geq 0$
$4b^4ca + 6b^3c^2a + 6b^2c^3a + 4bc^4a - b^4c^2 - 2b^3c^3 - c^4b^2\geq 2b^3c^2a + 2b^2c^3a - b^4c^2 - 2b^3c^3 - c^4b^2\geq 0$

so..............

$\frac {3(ab + bc + ca)}{2(a^2b^2 + b^2c^2 + c^2a^2)}\le\frac1{a^2 + bc} + \frac1{b^2 + ca} + \frac1{c^2 + ab}$

is equivalent to

$((2b^3 - b^2c - c^2b + 2c^3)a^3 + ( - b^2c^2 + b^4 + c^4)a^2 + ( - 2b^4c - 2c^4b + 3c^5 + 3b^5)a$
$- c^5b - b^5c - b^4c^2 - c^4b^2 + b^3c^3 + 3c^6 + 3b^6)(a - b)(a - c) + (b + c)((3a - 3c)b^4$
$+ 2c(a - c)b^3 + 2c^2(2a - c)b^2 + c^3( - 3c + 2a)b + 3c^4a)(b - c)^2\geq 0$

$2b^3 - b^2c - c^2b + 2c^3\geq b^3+c^3-c^2b-b^2c=(b+c)(b-c)^2\geq 0$
$- b^2c^2 + b^4 + c^4\geq 0$
$- 2b^4c - 2c^4b + 3c^5 + 3b^5\geq - 2b^4c - 2c^4b + 2c^5 + 2b^5=2(c+b)(b^2+c^2)(-c+b)^2\geq 0$
$- c^5b - b^5c - b^4c^2 - c^4b^2 + b^3c^3 + 3c^6 + 3b^6$
$\geq (b^6+c^6-c^5b-b^5c)+(b^6+c^6-b^4c^2-c^4b^2)$
$=(b^4+b^3c+b^2c^2+c^3b+c^4)(-c+b)^2+(b^2+c^2)(-c+b)^2(c+b)^2\geq 0$

$(3a - 3c)b^4 + 2c(a - c)b^3 + 2c^2(2a - c)b^2 + c^3( - 3c + 2a)b + 3c^4a\geq 0$

so..........

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arqady

#3 h Nov 11, 2009, 4:07 PM • 2 Y

Y by Adventure10, Mango247

dduclam wrote:

For all $a,b,c$ be nonnegative real numbers, show that
$\frac {3(ab + bc + ca)}{2(a^2b^2 + b^2c^2 + c^2a^2)}\le\frac1{a^2 + bc} + \frac1{b^2 + ca} + \frac1{c^2 + ab}\le\frac {3(a^2 + b^2 + c^2)}{2(a^2b^2 + b^2c^2 + c^2a^2)}$

Let $a+b+c=3u,$ $ab+ac+bc=3v^2$ and $abc=w^3.$
Hence, the left inequality is equivalent to $f(w^3)\geq0,$ where $f$ is decreasing function.
Thus, for the proof enough to check only one case: $b=c=1,$ which gives $(a-1)^2(2a^2-a+3)\geq0.$
The right inequality is equivalent to $g(w^3)\geq0,$ where $f$ is increasing function.
Thus, here enough to check two cases:
1) $b=1$ and $c=0$ ;
2) $b=c=1.$

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#4 h Nov 11, 2009, 5:24 PM • 2 Y

Y by Adventure10, Mango247

I can't do the case $a = b = c = 0$ . Is there something wrong in the problem?

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arqady

#5 h Nov 11, 2009, 6:13 PM • 2 Y

Y by Adventure10, Mango247

MeKnowsNothing wrote:

I can't do the case $a = b = c = 0$ . Is there something wrong in the problem?

It means $a,$ $b$ and $c$ are non-negative numbers such that $ab+ac+bc\neq0.$

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#6 h Nov 12, 2009, 12:29 PM • 2 Y

Y by Adventure10, Mango247

dduclam wrote:

For all $a,b,c$ be nonnegative real numbers, show that
$\frac {3(ab + bc + ca)}{2(a^2b^2 + b^2c^2 + c^2a^2)}\le\frac1{a^2 + bc} + \frac1{b^2 + ca} + \frac1{c^2 + ab}\le\frac {3(a^2 + b^2 + c^2)}{2(a^2b^2 + b^2c^2 + c^2a^2)}$

I think the following inequality is true also .

Let $a,b,c$ be nonnegative real numbers .

Prove that :
$\ \frac {1}{a^2 + 2bc} + \frac {1}{b^2 + 2ca} + \frac {1}{c^2 + 2ab} \geq \frac {ab + bc + ca}{a^2b^2 + b^2c^2 + c^2a^2}$

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Vincent Gilbert

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Vincent Gilbert

#7 h Nov 12, 2009, 2:07 PM • 2 Y

Y by Adventure10, Mango247

$(\sum a^2b^2)( \sum \frac{1}{a^2+bc} ) \le \frac{3}{2}(\sum a^2)$
$(\sum a^2)( \sum \frac{ a^2}{a^2+bc})-\sum a^2+\sum bc \le \frac{3}{2}(\sum a^2)$
$\frac{1}{2}.( \sum a)^2 \le (\sum a^2)( \sum \frac{bc}{a^2+bc})$
We have :
$\sum \frac{bc}{a^2+bc} \ge \frac{ (\sum ab)^2}{ \sum a^2bc+\sum b^2c^2} \ge \frac{ (\sum a)^2}{2(\sum a^2)}$

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#8 h Nov 12, 2009, 3:04 PM • 1 Y

Y by Adventure10

For elementary proof...
$\sum_{cyc} \frac {1}{a^2 + bc}\geq \frac {3(ab + bc + ca)}{2(a^2b^2 + b^2c^2 + c^2a^2)}$
from cauchy
$\sum_{cyc} \frac {(b + c)^2}{(a^2 + bc)(b + c)^2}\geq \frac {4(\sum_{cyc} a)^2}{\sum_{sym} a^3b + 2\sum_{cyc} a^2bc + 4\sum_{cyc} a^2b^2}\geq \frac {3(ab + bc + ca)}{2(a^2b^2 + b^2c^2 + c^2a^2)}$
which is equivalent to
$8\sum_{sym} a^4b^2 + 24a^2b^2c^2 + 16\sum_{cyc} a^3b^3 + 16\sum_{sym} a^3b^2c\geq 3\sum_{sym} a^4b^2 + 21\sum_{sym} a^3b^2c + 6\sum_{cyc} a^4bc + 18a^2b^2c^2 + 12\sum_{cyc} a^3b^3$
or
$5\sum_{sym} a^4b^2 + 6a^2b^2c^2 + 4\sum_{cyc} a^3b^3\geq 5\sum_{sym} a^3b^2c + 6\sum_{cyc} a^4bc$
which is obviously true from schur and muirhead... ( $\sum_{cyc} a^3b^3 + 3a^2b^2c^2\geq \sum_{sym} a^3b^2c$ )

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#9 h Nov 18, 2009, 7:13 PM • 2 Y

Y by Adventure10, Mango247

Vincent Gilbert wrote:

$(\sum a^2b^2)( \sum \frac {1}{a^2 + bc} ) \le \frac {3}{2}(\sum a^2)$
$(\sum a^2)( \sum \frac { a^2}{a^2 + bc}) - \sum a^2 + \sum bc \le \frac {3}{2}(\sum a^2)$
$\frac {1}{2}.( \sum a)^2 \le (\sum a^2)( \sum \frac {bc}{a^2 + bc})$
We have :
$\sum \frac {bc}{a^2 + bc} \ge \frac { (\sum ab)^2}{\sum a^2bc + \sum b^2c^2} \ge \frac { (\sum a)^2}{2(\sum a^2)}$

Nice, Vincent Gilbert, the first part of your solution is similar to me. After that, I use the lemma: $\sum_{cyc}\frac {bc(b^2 + c^2)}{a^2 + bc}\ge a^2 + b^2 + c^2$ to completed the proof.

Rose_joker wrote:

For elementary proof...
$\sum_{cyc} \frac {1}{a^2 + bc}\geq \frac {3(ab + bc + ca)}{2(a^2b^2 + b^2c^2 + c^2a^2)}$
from cauchy
$\sum_{cyc} \frac {(b + c)^2}{(a^2 + bc)(b + c)^2}\geq \frac {4(\sum_{cyc} a)^2}{\sum_{sym} a^3b + 2\sum_{cyc} a^2bc + 4\sum_{cyc} a^2b^2}\geq \frac {3(ab + bc + ca)}{2(a^2b^2 + b^2c^2 + c^2a^2)}$
which is equivalent to
$8\sum_{sym} a^4b^2 + 24a^2b^2c^2 + 16\sum_{cyc} a^3b^3 + 16\sum_{sym} a^3b^2c\geq 3\sum_{sym} a^4b^2 + 21\sum_{sym} a^3b^2c + 6\sum_{cyc} a^4bc + 18a^2b^2c^2 + 12\sum_{cyc} a^3b^3$
or
$5\sum_{sym} a^4b^2 + 6a^2b^2c^2 + 4\sum_{cyc} a^3b^3\geq 5\sum_{sym} a^3b^2c + 6\sum_{cyc} a^4bc$
which is obviously true from schur and muirhead... ( $\sum_{cyc} a^3b^3 + 3a^2b^2c^2\geq \sum_{sym} a^3b^2c$ )

You are right, Rose_joker. But we have a shorter proof only with Cauchy-Schwarz.

Materazzi wrote:

dduclam wrote:

For all $a,b,c$ be nonnegative real numbers, show that
$\frac {3(ab + bc + ca)}{2(a^2b^2 + b^2c^2 + c^2a^2)}\le\frac1{a^2 + bc} + \frac1{b^2 + ca} + \frac1{c^2 + ab}\le\frac {3(a^2 + b^2 + c^2)}{2(a^2b^2 + b^2c^2 + c^2a^2)}$

I think the following inequality is true also .

Let $a,b,c$ be nonnegative real numbers .

Prove that :
$\ \frac {1}{a^2 + 2bc} + \frac {1}{b^2 + 2ca} + \frac {1}{c^2 + 2ab} \geq \frac {ab + bc + ca}{a^2b^2 + b^2c^2 + c^2a^2}$

You are right, Materazzi. Infact, we have the following inequality for all $a,b,c$ be nonnegative relnumbers
$\sum_{cyc}\frac1{a^2 + bc}\ge\sum_{cyc}\frac1{a^2 + 2bc} + \frac {ab + bc + ca}{2(a^2b^2 + b^2c^2 + c^2a^2)}$

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manlio

#10 h Nov 19, 2009, 12:43 PM • 1 Y

Y by Adventure10

Dear dduclam,

I proved $\sum_{cyc}\frac {bc(b^2 + c^2)}{a^2 + bc}\ge a^2 + b^2 + c^2$

this way: By BCS $LHS \geq \frac{(\sum ab\sqrt{a^2+b^2})^2}{\sum bc(a^2+bc)}$

so it suffices to prove:

$(\sum bc\sqrt{b^2+c^2})^2 \geq (\sum a^2)(abc\sum a+\sum a^2b^2)$

Opening LHS and using $\sqrt{a^2+b^2}{\sqrt{a^2+c^2} \geq a^2 +bc}$

we get shur of degree 3. Using this LEMMA to prove your inequality I obtained:

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

that I cannot prove.

Can you please help me to end your proof?

Thank you very much.

Can you also post your proof for second inequality by only Cauchy? I am very intereste in it. Thank you very much again for thes nice inequalities.

dduclam wrote:

Vincent Gilbert wrote:

$(\sum a^2b^2)( \sum \frac {1}{a^2 + bc} ) \le \frac {3}{2}(\sum a^2)$
$(\sum a^2)( \sum \frac { a^2}{a^2 + bc}) - \sum a^2 + \sum bc \le \frac {3}{2}(\sum a^2)$
$\frac {1}{2}.( \sum a)^2 \le (\sum a^2)( \sum \frac {bc}{a^2 + bc})$
We have :
$\sum \frac {bc}{a^2 + bc} \ge \frac { (\sum ab)^2}{\sum a^2bc + \sum b^2c^2} \ge \frac { (\sum a)^2}{2(\sum a^2)}$

Nice, Vincent Gilbert, the first part of your solution is similar to me. After that, I use the lemma: $\sum_{cyc}\frac {bc(b^2 + c^2)}{a^2 + bc}\ge a^2 + b^2 + c^2$ to completed the proof.

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#11 h Nov 19, 2009, 5:45 PM • 2 Y

Y by Adventure10, Mango247

manlio wrote:

Dear dduclam,

I proved $\sum_{cyc}\frac {bc(b^2 + c^2)}{a^2 + bc}\ge a^2 + b^2 + c^2$

this way: By BCS $LHS \geq \frac {(\sum ab\sqrt {a^2 + b^2})^2}{\sum bc(a^2 + bc)}$

so it suffices to prove:

$(\sum bc\sqrt {b^2 + c^2})^2 \geq (\sum a^2)(abc\sum a + \sum a^2b^2)$

Opening LHS and using $\sqrt {a^2 + b^2}{\sqrt {a^2 + c^2} \geq a^2 + bc}$

we get shur of degree 3. Using this LEMMA to prove your inequality I obtained:

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

that I cannot prove.

Can you please help me to end your proof?

Thank you very much.

Very nice your proof, manlio

Using this lemma to prove my Inequality as follow

We have
$(a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})\left(\sum_{cyc}\frac {1}{a^{2} + bc}\right) = \sum_{cyc}\left(b^2 + c^2 - bc\frac {b^2 + c^2 - bc}{a^2 + bc}\right)$
The Inequality is equivalent to
$2(a^2 + b^2 + c^2) - \sum_{cyc}bc\frac {b^2 + c^2 - bc}{a^2 + bc}\le\frac3{2}(a^2 + b^2 + c^2)$
or
$2\sum_{cyc}bc\frac {b^2 + c^2 - bc}{a^2 + bc}\ge a^2 + b^2 + c^2$
By AM-GM Inequality, we have $2(b^2 + c^2 - bc)\ge b^2 + c^2$ , then using above lemma, we have done!

manlio wrote:

Can you also post your proof for second inequality by only Cauchy? I am very intereste in it. Thank you very much again for thes nice inequalities.

The proof for second Inequality. We have
$(a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})\left(\sum_{cyc}\frac {1}{a^{2} + bc}\right) = \sum_{cyc}\left(bc + \frac {a^2(b^2 + c^2 - bc)}{a^2 + bc}\right)$
The Inequality is equivalent to
$2\sum_{cyc}\frac {a^2(b^2 - bc + c^2)}{a^2 + bc}\ge ab + bc + ca$
or
$2\sum_{cyc}a^2\left(1+\frac {b^2 - bc + c^2}{a^2 + bc} \right)\ge 2(a^2 + b^2 + c^2)+ab + bc + ca$
or
$\sum_{cyc}\frac {a^2}{a^2 + bc}\ge1 + \frac {ab + bc + ca}{2(a^2 + b^2 + c^2)}$
By Cauchy-Schwarz Inequality, we get
$\sum_{cyc}\frac {a^2}{a^2 + bc}\ge\frac {(a + b + c)^2}{\sum_{cyc}a^2 + \sum_{cyc}bc} = 1 + \frac {ab + bc + ca}{\sum_{cyc}a^2 + \sum_{cyc}bc}\ge 1 + \frac {ab + bc + ca}{2(a^2 + b^2 + c^2)}$
We completed the proof.

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manlio

#12 h Nov 19, 2009, 8:02 PM • 2 Y

Y by Adventure10, Mango247

Very nice your proof, dduclam.

Thank you very much.

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manlio

#13 h Nov 20, 2009, 6:53 AM • 2 Y

Y by Adventure10, Mango247

Dear dduclam,

did you prove these very nice inequalities by SOS or have you a neat proof by Cauchy-Schwarz?

Thank you very much.

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

$\sum_{cyc}\frac1{a^2 + bc}\ge\sum_{cyc}\frac1{a^2 + 2bc} + \frac {ab + bc + ca}{2(a^2b^2 + b^2c^2 + c^2a^2)}$

For the first inequality I found this solution on VIMF:
http://ddbdt.co.cc/forum/showthread.php?t=238

It is nice but it uses SOS method, so I would like to know if you have a solution by only BCS.

Thank you very much

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#14 h Nov 21, 2009, 4:52 AM • 2 Y

Y by Adventure10, Mango247

manlio wrote:

Dear dduclam,

did you prove these very nice inequalities by SOS or have you a neat proof by Cauchy-Schwarz?

Thank you very much.

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

$\sum_{cyc}\frac1{a^2 + bc}\ge\sum_{cyc}\frac1{a^2 + 2bc} + \frac {ab + bc + ca}{2(a^2b^2 + b^2c^2 + c^2a^2)}$
For the first inequality I found this solution on VIMF:
http://ddbdt.co.cc/forum/showthread.php?t=238

It is nice but it uses SOS method, so I would like to know if you have a solution by only BCS.

Thank you very much

Hint for you, manlio:
$2\sum_{cyc}\frac{a^2(b^2-bc+c^2)}{a^2+bc}\ge\sum_{cyc}\frac{a^2(2b^2-bc+2c^2)}{a^2+2bc}\ge ab+bc+ca$

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manlio

#15 h Nov 22, 2009, 7:00 PM • 2 Y

Y by Adventure10, Mango247

Dear dduclam,

thank you very much for your hint.

$2\sum_{cyc}\frac {a^2(b^2 - bc + c^2)}{a^2 + bc}\ge\sum_{cyc}\frac {a^2(2b^2 - bc + 2c^2)}{a^2 + 2bc}\ge ab + bc + ca$

$1)$ $2\sum_{cyc}\frac {a^2(b^2 - bc + c^2)}{a^2 + bc}\ge\sum_{cyc}\frac {a^2(2b^2 - bc + 2c^2)}{a^2 + 2bc}$

$2)$ $\sum_{cyc}\frac {a^2(2b^2 - bc + 2c^2)}{a^2 + 2bc}\ge ab + bc + ca$

I tried the first inequality, but I failed

For the second I wrote it as

$2\sum a^2 \sum \frac{a^2}{a^2+2bc} +3abc \sum\frac{a}{a^2+2bc} \geq \sum ab+2\sum a^2$

But using $\sum \frac{a^2}{a^2+2bc} \geq 1$ it is too much weak to prove it.

Can you please help me?

I don't understand how your hint can be used to prove your inequalities?

Thank you very much.

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#16 h Nov 27, 2009, 7:47 PM • 1 Y

Y by Adventure10

arqady wrote:

Hence, the left inequality is equivalent to $f(w^3)\geq0,$ where $f$ is decreasing function.
The right inequality is equivalent to $g(w^3)\geq0,$ where $f$ is increasing function.

Can you post the whole form of $f$ and $g$ , or the only thing you need to know is that it is a function of $w$ ,whatever is it .

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arqady

#17 h Nov 28, 2009, 9:56 AM • 2 Y

Y by Adventure10, Mango247

enndb0x wrote:

arqady wrote:

Hence, the left inequality is equivalent to $f(w^3)\geq0,$ where $f$ is decreasing function.
The right inequality is equivalent to $g(w^3)\geq0,$ where $f$ is increasing function.

Can you post the whole form of $f$ and $g$

Certainly!
$f(w^3)=8(u^2-v^2)w^6-(63u^3v^2-54uv^4)w^3+54u^2v^6-45v^8.$
$g(w^3)=16(u^2-v^2)w^6+(81u^5-180u^3v^2+108uv^4)w^3+27u^2v^6-36v^8.$

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#18 h Jan 25, 2010, 6:56 AM • 2 Y

Y by Adventure10, Mango247

manlio wrote:

Dear dduclam,

thank you very much for your hint.
$2\sum_{cyc}\frac {a^2(b^2 - bc + c^2)}{a^2 + bc}\ge\sum_{cyc}\frac {a^2(2b^2 - bc + 2c^2)}{a^2 + 2bc}\ge ab + bc + ca$
$1)$ $2\sum_{cyc}\frac {a^2(b^2 - bc + c^2)}{a^2 + bc}\ge\sum_{cyc}\frac {a^2(2b^2 - bc + 2c^2)}{a^2 + 2bc}$

$2)$ $\sum_{cyc}\frac {a^2(2b^2 - bc + 2c^2)}{a^2 + 2bc}\ge ab + bc + ca$

I tried the first inequality, but I failed

For the second I wrote it as

$2\sum a^2 \sum \frac {a^2}{a^2 + 2bc} + 3abc \sum\frac {a}{a^2 + 2bc} \geq \sum ab + 2\sum a^2$

But using $\sum \frac {a^2}{a^2 + 2bc} \geq 1$ it is too much weak to prove it.

Can you please help me?

I don't understand how your hint can be used to prove your inequalities?

Thank you very much.

Dear Manlio, I will explain carefully about my post.
The first Inequality,
$\ \frac {1}{a^{2} + 2bc} + \frac {1}{b^{2} + 2ca} + \frac {1}{c^{2} + 2ab}\geq\frac {ab + bc + ca}{a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2}}$
we use the equality $2(a^2b^2 + b^2c^2 + c^2a^2) = bc(a^2 + 2bc) + a^2(2b^2 + 2c^2 - a^2)$ and rewrite the inequality as
$\sum_{cyc}\frac {a^2(2b^2 - bc + 2c^2)}{a^2 + 2bc}\ge ab + bc + ca$
or
$\left(2\sum_{cyc}\frac {a^2(2b^2 - bc + 2c^2)}{a^2 + 2bc} + a^2\right)\ge (a + b + c)^2$
or
$\sum_{cyc}\frac {a^2(a^2 + 4b^2 + 4c^2)}{a^2 + 2bc}\ge (a + b + c)^2$
By Cauchy-Chwarz Inequality, we have
$\sum_{cyc}\frac {a^2(a^2 + 4b^2 + 4c^2)}{a^2 + 2bc}\ge \frac {\left(\sum_{cyc}a\sqrt {a^2 + 4b^2 + 4c^2}\right)^2}{(a + b + c)^2}$
Therefore, it suffices to prove that
$\sum_{cyc}a\sqrt {a^2 + 4b^2 + 4c^2}\ge (a + b + c)^2$
This Inequality was proven at here.

The second Inequality
$\sum_{cyc}\frac {1}{a^{2} + bc}\ge\sum_{cyc}\frac {1}{a^{2} + 2bc} + \frac {ab + bc + ca}{2(a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})}$
by also the equality $2(a^2b^2 + b^2c^2 + c^2a^2) = bc(a^2 + 2bc) + a^2(2b^2 + 2c^2 - a^2)$ and $a^2b^2 + b^2c^2 + c^2a^2 = bc(a^2 + bc) + a^2(b^2 + c^2 - bc)$ , we rewrite the inequality as
$2\sum_{cyc}\frac {a^2(b^2 - bc + c^2)}{a^2 + bc}\ge\sum_{cyc}\frac {a^2(2b^2 - bc + 2c^2)}{a^2 + 2bc}$
or
$2\sum\frac {b^2 - bc + c^2}{1 + x}\ge\sum\frac {2b^2 - bc + 2c^2}{1 + 2x}$
where $x = \frac {bc}{a^2}, y = \frac {ca}{b^2}, z = \frac {ab}{c^2}.$
After that, we use expansion to complete the proof, but this way is not meaningful (althought the first Inequality is also use this way).

This post has been edited 2 times. Last edited by dduclam, Jan 25, 2010, 5:02 PM

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manlio

#19 h Jan 25, 2010, 4:22 PM • 2 Y

Y by Adventure10, Mango247

Thank you very much, dduclam. I will study your solution very carefully.

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#20 h Jan 27, 2010, 12:10 PM • 2 Y

Y by Adventure10, Mango247

I tried to solve this

$2\sum\frac {b^2 - bc + c^2}{1 + x}\ge\sum\frac {2b^2 - bc + 2c^2}{1 + 2x}$
where $x = \frac {bc}{a^2}, y = \frac {ca}{b^2}, z = \frac {ab}{c^2}.$

by full expansion, but it is terribly long.

Has anyone a trick to simplify it, please? Thank you very much.

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#21 h Feb 1, 2016, 11:53 AM • 1 Y

Y by Adventure10

Materazzi wrote:

Let $a,b,c$ be nonnegative real numbers. Prove that
$\ \frac {1}{a^2 + 2bc} + \frac {1}{b^2 + 2ca} + \frac {1}{c^2 + 2ab} \geq \frac {ab + bc + ca}{a^2b^2 + b^2c^2 + c^2a^2} \quad (1)$

Put $p=a+b+c,\,q=ab+bc+ca$ and $r=abc$ we have lemma
$2p(2p^2-3q)r \leqslant q^2(p^2-q). \quad (2)$ Write inequality $(1)$ as
$\frac{\displaystyle \sum (a^2+2bc)(b^2+2ca)}{(a^2+2bc)(b^2+2ca)(c^2+2ab)} \geqslant \frac{ab+bc+ca}{a^2b^2+b^2c^2+c^2a^2},$ equivalent to
$\frac{q(2p^2-3q)}{27r^2+2p(2p^2-9q)r+2q^3} \geqslant \frac{q}{q^2-2pr},$ or
$q(2p^2-3q)(q^2-2pr) \geqslant q\left[27r^2+2p(2p^2-9q)r+2q^3\right],$ $\left[q^2(p^2-q)-2p(2p^2-3q)r\right]+\left[p^2q^2-4q^3+2p(9q-2p^2)r-27r^2\right] \geqslant 0.$ Whicch is true because lemma $(2)$ and
$p^2q^2-4q^3+2p(9q-2p^2)r-27r^2=(a-b)^2(b-c)^2(c-a)^2 \ge 0.$ We are done.

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