A nice inequality!!

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Michael

99 posts

Michael

#1 h Mar 30, 2006, 6:57 PM • 6 Y

Y by rightways, adityaguharoy, Adventure10, HWenslawski, Mango247, Marcus_Zhang

Hello!

Let $a,b,c$ be three real numbers such that $abc=1$ . Prove that:
${{1}\over{a^2+ab}}+{{1}\over{b^2+bc}}+{{1}\over{c^2+ca}}\geq {3\over 2}.$

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arqady

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arqady

#2 h Mar 31, 2006, 6:17 PM • 3 Y

Y by rightways, Adventure10, HWenslawski

Michael wrote:

Hello!

Let $a,b,c$ be three real numbers such that $abc=1$ . Prove that:
${{1}\over{a^2+ab}}+{{1}\over{b^2+bc}}+{{1}\over{c^2+ca}}\geq {3\over 2}.$

I think $a,b,c$ are positive real numbers.

Let $a=\frac{y}{x},$ $b=\frac{z}{y},$ $c=\frac{x}{z}.$
Hence, $\sum_{cyc}\frac{1}{a^2+ab}\geq\frac{3}{2}\Leftrightarrow\sum_{cyc}\frac{x^2}{y^2+xz}\geq\frac{3}{2}.$
$\sum_{cyc}\frac{x^2}{y^2+xz}=\sum_{cyc}\frac{x^4}{x^2y^2+x^3z}\geq\frac{(x^2+y^2+z^2)^2}{x^2y^2+x^2z^2+y^2z^2+x^3z+y^3x+z^3y}.$
But $\frac{(x^2+y^2+z^2)^2}{x^2y^2+x^2z^2+y^2z^2+x^3z+y^3x+z^3y}\geq\frac{3}{2}\Leftrightarrow\sum_{cyc}(2x^4+x^2y^2-3x^3z)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(8x^4+4x^2z^2-12x^3z)\geq0\Leftrightarrow\sum_{cyc}(7x^4+z^4+4x^2z^2-12x^3z)\geq0.$
But $7x^4+z^4+4x^2z^2\geq12\cdot\sqrt[12]{x^{28+8}z^{4+8}}=12x^3z.$

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Michael

99 posts

Michael

#3 h Apr 1, 2006, 10:57 AM • 4 Y

Y by Adventure10, Adventure10, Mango247, Marcus_Zhang

Thank you very much arqady. Yes, you are right, the real numbers $a,b,c$ are positive and verify $abc=1$ .

Does your solution apply also for more than three variables (for examples for four real positive numbers $a,b,c,d$ such that $abcd=1$ , or more generally for $n$ reals). I think that in this case the RHS of the inequality should be $n/2$ .

Michael.

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arqady

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arqady

#4 h Apr 1, 2006, 8:42 PM • 2 Y

Y by Adventure10, Mango247

Michael wrote:

I think that in this case the RHS of the inequality should be $n/2$ .
Michael.

This is very nice!

I had difficulties already for n=5.

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Michael

99 posts

Michael

#5 h Apr 2, 2006, 6:26 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

Michael wrote:

I think that in this case the RHS of the inequality should be $n/2$ .
Michael.

This is very nice!

I had difficulties already for n=5.

Did you solve it for $n=4$ ?

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arqady

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arqady

#6 h Apr 2, 2006, 10:45 AM • 1 Y

Y by Adventure10

Michael wrote:

Did you solve it for $n=4$ ?

Yes. Perheps to prove it like for the case $n=3.$

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Vasc

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Vasc

#7 h Aug 8, 2006, 8:03 AM • 2 Y

Y by Adventure10, Mango247

The more general statement is valid:
Let $p$ and $a,b,c$ be positive numbers such that $abc=1$ . Prove that:
${{1}\over{a(a+pb)}}+{{1}\over{b(b+pc)}}+{{1}\over{c(c+pa)}}\geq{3\over{1+p}}$ .

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arqady

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arqady

#8 h Jan 4, 2011, 9:34 AM • 3 Y

Y by kprepaf, Adventure10, Mango247

Vasc wrote:

The more general statement is valid:
Let $p$ and $a,b,c$ be positive numbers such that $abc=1$ . Prove that:
${{1}\over{a(a+pb)}}+{{1}\over{b(b+pc)}}+{{1}\over{c(c+pa)}}\geq{3\over{1+p}}$ .

For the proof we can use the following your inequality.
$(x^2+y^2+z^2)^2\geq3(x^3y+y^3z+z^3x)$

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Unknown-6174

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Unknown-6174

#9 h Sep 24, 2013, 5:07 PM • 2 Y

Y by Adventure10 and 1 other user

Michael wrote:

Hello!

Let $a,b,c$ be three real numbers such that $abc=1$ . Prove that:
${{1}\over{a^2+ab}}+{{1}\over{b^2+bc}}+{{1}\over{c^2+ca}}\geq {3\over 2}.$

An easier method is using AM-HM.
$(\frac{1}{a(a+b)} +\frac{1}{b(b+c)} +\frac{1}{c(c+a)})(a(a+b) + b(b+c) + c(c+a)) \ge 9$
$a^2 +ab +b^2 +bc +c^2 +ac \ge 6$
From which the result follows

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123steeve

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123steeve

#10 h Sep 24, 2013, 5:29 PM • 3 Y

Y by MathArt4, Adventure10, Mango247

Unknown-6174 wrote:

An easier method is using AM-HM.
$(\frac{1}{a(a+b)} +\frac{1}{b(b+c)} +\frac{1}{c(c+a)})(a(a+b) + b(b+c) + c(c+a)) \ge 9$
$a^2 +ab +b^2 +bc +c^2 +ac \ge 6$
From which the result follows

Doesn't work - need $a^2 +ab +b^2 +bc +c^2 +ac \le 6$

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Unknown-6174

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Unknown-6174

#11 h Oct 19, 2013, 12:44 PM • 1 Y

Y by Adventure10

@123steeve:Sorry my bad!

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Nguyenngoctu

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Nguyenngoctu

#12 h May 1, 2016, 10:01 AM • 1 Y

Y by Adventure10

arqady wrote:

Michael wrote:

Hello!

Let $a,b,c$ be three real numbers such that $abc=1$ . Prove that:
${{1}\over{a^2+ab}}+{{1}\over{b^2+bc}}+{{1}\over{c^2+ca}}\geq {3\over 2}.$

I think $a,b,c$ are positive real numbers.

Let $a=\frac{y}{x},$ $b=\frac{z}{y},$ $c=\frac{x}{z}.$
Hence, $\sum_{cyc}\frac{1}{a^2+ab}\geq\frac{3}{2}\Leftrightarrow\sum_{cyc}\frac{x^2}{y^2+xz}\geq\frac{3}{2}.$
$\sum_{cyc}\frac{x^2}{y^2+xz}=\sum_{cyc}\frac{x^4}{x^2y^2+x^3z}\geq\frac{(x^2+y^2+z^2)^2}{x^2y^2+x^2z^2+y^2z^2+x^3z+y^3x+z^3y}.$
But $\frac{(x^2+y^2+z^2)^2}{x^2y^2+x^2z^2+y^2z^2+x^3z+y^3x+z^3y}\geq\frac{3}{2}\Leftrightarrow\sum_{cyc}(2x^4+x^2y^2-3x^3z)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(8x^4+4x^2z^2-12x^3z)\geq0\Leftrightarrow\sum_{cyc}(7x^4+z^4+4x^2z^2-12x^3z)\geq0.$
But $7x^4+z^4+4x^2z^2\geq12\cdot\sqrt[12]{x^{28+8}z^{4+8}}=12x^3z.$

Prove: $2\left( {{x^4} + {y^4} + {z^4}} \right) + {x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2} \ge 3\left( {x{y^3} + y{z^3} + z{x^3}} \right)$
We have ${x^4} + {x^2}{z^2} \ge 2z{x^3}$ . Similarly $\Rightarrow {x^4} + {y^4} + {z^4} + {x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2} \ge 2\left( {x{y^3} + y{z^3} + z{x^3}} \right)$ .
And ${x^4} + {y^4} + {z^4} \ge x{y^3} + y{z^3} + z{x^3}$ by Muirhead inequality: $\left[ {4,0,0} \right] \ge \left[ {3,1,0} \right]$ .
It's true or false?

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arqady

30147 posts

arqady

#13 h May 1, 2016, 7:52 PM • 2 Y

Y by Adventure10, Mango247

I think your last step is not so good.

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bobthesmartypants

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bobthesmartypants

#14 h May 1, 2016, 8:04 PM • 2 Y

Y by Adventure10, Mango247

$x^4+y^4+z^4\ge xy^3+yz^3+zx^3$ is true by Rearrangement though so it is an easy fix.

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Nguyenngoctu

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Nguyenngoctu

#15 h May 1, 2016, 10:34 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

I think your last step is not so good.

Dear arquady! Why? Can you tell me more okay?

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sqing

40998 posts

sqing

#16 h Dec 15, 2017, 2:59 AM • 2 Y

Y by Adventure10, Mango247

Michael wrote:

Hello!
Let $a,b,c$ be three real numbers such that $abc=1$ . Prove that:
${{1}\over{a^2+ab}}+{{1}\over{b^2+bc}}+{{1}\over{c^2+ca}}\geq {3\over 2}.$

Let $a,b,c$ be three real numbers such that $abc=1$ . Prove that:
$\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{2}.$
Zhautykov Olympiad, Kazakhstan 2008

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viki2001

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viki2001

#17 h Dec 15, 2017, 5:03 AM • 2 Y

Y by Adventure10, Mango247

very nice

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e61442289

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e61442289

#18 h Jul 30, 2021, 8:29 PM

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I have a solution (I think) with MA - MH

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sqing

40998 posts

sqing

#19 h Sep 16, 2021, 1:44 PM

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sqing wrote:

Let $a,b,c$ be three real numbers such that $abc=1$ . Prove that:
$\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{2}.$ Zhautykov Olympiad, Kazakhstan 2008

Let $a,b,c$ be positive real numbers such that $abc=1$ . Prove that $\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq\frac{3}{2}.$

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sqing

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sqing

#20 h Sep 16, 2021, 2:01 PM

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Let $a,b,c$ be positive real numbers such that $abc=1$ . Prove that $\frac{2}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq2\sqrt 2-1.$ Equality holds when $a=\sqrt 2+1,b=1,c=\sqrt 2-1.$
$\frac{3}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq2\sqrt 3-\frac{3}{2}.$ Equality holds when $a=2\sqrt 3+3,b=1,c=\frac{2}{\sqrt 3}-1.$
$\frac{k}{a(b+1)}+\frac{k}{b(c+1)}+\frac{1}{c(a+1)}\geq2\sqrt k-\frac{1}{2}.$ Where $k> \frac{1}{4}.$ Equality holds when $a=1,b=2\sqrt k-1,c=\frac{1}{2\sqrt k-1}.$

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sqing

40998 posts

sqing

#21 h Jun 21, 2022, 1:44 PM

Y by

sqing wrote:

Let $a,b,c$ be three real numbers such that $abc=1$ . Prove that:
$\frac{1}{a(a+1)}+\frac{1}{b(b+1)}+\frac{1}{c(c+1)}\ge \frac{3}{2}.$

Let $a, b, c$ be three real numbers such that $abc = 1$ . Prove that $\frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)} \geq \frac{3}{4}$ S
h

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sqing

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sqing

#22 h Jun 22, 2022, 7:16 AM

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sqing wrote:

Let $a, b, c$ be three real numbers such that $abc = 1$ . Prove that $\frac{a}{(b+1)(c+1)}+\frac{b}{(c+1)(a+1)}+\frac{c}{(a+1)(b+1)} \geq \frac{3}{4}$

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