Ineq 27 (nice)

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math10

478 posts

math10

#1 h Nov 20, 2009, 10:41 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be positive numbers such that $abc=1$ . Prove that:
$a^2+b^2+c^2+9(ab+ac+bc) \geq 10(a+b+c)$

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arqady

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arqady

#2 h Nov 20, 2009, 10:53 AM • 2 Y

Y by Adventure10, Mango247

math10 wrote:

Let $a,b,c$ be positive numbers such that $abc = 1$ . Prove that:
$a^2 + b^2 + c^2 + 9(ab + ac + bc) \geq 10(a + b + c)$

Let $a + b + c = 3u,$ $ab + ac + bc = 3v^2$ and $abc = w^3.$
Hence, $\sum_{cyc}(a^2 + 9ab - 10a)\geq0\Leftrightarrow w^3\leq\left(\frac {3u^2 + 7v^2}{10u}\right)^3.$
But $w^3$ gets a maximal value when two numbers from $\{a,b,c\}$ are equal.
Thus, it remains to prove your inequality for $b = c = t > 0$ and $a = \frac {1}{t^2},$ which is very easy.
By the way the following inequality is also true.
Let $a,$ $b$ and $c$ are positive numbers such that $abc = 1.$ Prove that:

$a^2 + b^2 + c^2 + 15(ab + ac + bc) \geq 16(a + b + c)$

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math10

478 posts

math10

#3 h Nov 20, 2009, 11:16 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

math10 wrote:

Let $a,b,c$ be positive numbers such that $abc = 1$ . Prove that:
$a^2 + b^2 + c^2 + 9(ab + ac + bc) \geq 10(a + b + c)$

Let $a + b + c = 3u,$ $ab + ac + bc = 3v^2$ and $abc = w^3.$
Hence, $\sum_{cyc}(a^2 + 9ab - 10a)\geq0\Leftrightarrow w^3\leq\left(\frac {3u^2 + 7v^2}{10u}\right)^3.$
But $w^3$ gets a maximal value when two numbers from $\{a,b,c\}$ are equal.
Thus, it remains to prove your inequality for $b = c = t > 0$ and $a = \frac {1}{t^2},$ which is very easy.
By the way the following inequality is also true.
Let $a,$ $b$ and $c$ are positive numbers such that $abc = 1.$ Prove that:
$a^2 + b^2 + c^2 + 15(ab + ac + bc) \geq 16(a + b + c)$

I think that you had using method UVW(a method terrible than mix variable

)

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arqady

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arqady

#4 h Nov 20, 2009, 11:20 AM • 2 Y

Y by Adventure10, Mango247

math10 wrote:

I think that you had using method UVW(a method terrible than mix variable

)

Are you think that $uvw$ is a terrible method? I don't agree with you. It's a very strong method and easy to use it.

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math10

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math10

#5 h Nov 20, 2009, 1:25 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

math10 wrote:

I think that you had using method UVW(a method terrible than mix variable

)

Are you think that $uvw$ is a terrible method? I don't agree with you. It's a very strong method and easy to use it.

yes,it's really terrible, I've read the documentation about it but I just read exactly once

(complexity theory and need calculate many)

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enndb0x

843 posts

enndb0x

#6 h Nov 20, 2009, 6:35 PM • 2 Y

Y by Adventure10, Mango247

arqady : Why don't you write a detailed article about $uvw$ and post it on mathlinks , so everyone can understand it's beauty.I see you using that method very often , so I think you are the right person to write an article ,or maybe you're saving it for using in any of your books ? The posted article by Mathias DK is not very clear to me

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arqady

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arqady

#7 h Nov 20, 2009, 7:56 PM • 2 Y

Y by Adventure10, Mango247

enndb0x, I am ready to help you in the topic of Mathias DK.

By the way, you can ask all question about my proof here.

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enndb0x

843 posts

enndb0x

#8 h Nov 27, 2009, 9:05 PM • 2 Y

Y by Adventure10, Mango247

Thank you Arqady , to see if I understood I will try to solve the one you proposed.

Let $a + b + c = 3u ,ab + bc + ca = 3v^2$ and $abc = w^3$ .

Substituting these , the inequality can be transformed as $\left(\frac {3u^2 + 3v^2}{16u}\right)^3 \geq w^3$
But $w^3$ takes maximal value only when two of $\{ a,b,c \}$ are equal .In that case $a = b = t$ and $c = \frac {1}{t}$ ,the other part is calculations.

Can we just observe the function $f(v) = 3v^2 + 3u^2 - 16u$ (convex) ,and then to see when it takes minimal value ? Or maybe as a function of $u$ ,to see when it has minimum ?

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jgnr

1343 posts

jgnr

#9 h Nov 28, 2009, 5:33 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

math10 wrote:

$w^3\leq\left(\frac {3u^2 + 7v^2}{10u}\right)^3.$
But $w^3$ gets a maximal value when two numbers from $\{a,b,c\}$ are equal.

I don't understand this part. If we assume that two of $\{a,b,c\}$ are equal, wouldn't it affect the RHS too?

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arqady

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arqady

#10 h Nov 28, 2009, 11:17 AM • 2 Y

Y by Adventure10, Mango247

enndb0x wrote:

Thank you Arqady , to see if I understood I will try to solve the one you proposed.

Let $a + b + c = 3u ,ab + bc + ca = 3v^2$ and $abc = w^3$ .

Substituting these , the inequality can be transformed as $\left(\frac {3u^2 + 3v^2}{16u}\right)^3 \geq w^3$
But $w^3$ takes maximal value only when two of $\{ a,b,c \}$ are equal .In that case $a = b = t$ and $c = \frac {1}{t}$ ,the other part is calculations.

It's not true.

enndb0x wrote:

Can we just observe the function $f(v) = 3v^2 + 3u^2 - 16u$ (convex) ,and then to see when it takes minimal value ? Or maybe as a function of $u$ ,to see when it has minimum ?

We can. But it does not give noting. While if we need to prove that $f(u)\leq0$ it gives a case $b = c.$
By the way, if $f(v^2) = 3v^2 + 3u^2 - 16u$ then, $f$ is linear function.
Hence, $f$ gets a minimal value when two numbers from $\{a,b,c\}$ are equal.

Johan Gunardi wrote:

arqady wrote:

$w^3\leq\left(\frac {3u^2 + 7v^2}{10u}\right)^3.$
But $w^3$ gets a maximal value when two numbers from $\{a,b,c\}$ are equal.

I don't understand this part. If we assume that two of $\{a,b,c\}$ are equal, wouldn't it affect the RHS too?

We can assume that "RHS" is a constant.

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math10

478 posts

math10

#11 h Nov 28, 2009, 11:32 AM • 2 Y

Y by Adventure10, Mango247

I am waiting for the other nice solution.In topic,All people only using UVW.

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aakansh92

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aakansh92

#12 h Dec 1, 2009, 6:46 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

I noticed the following generalisation :
$\sum a^{2} +k \sum ab \geq (k+1) \sum a \ \forall\ k \in R$

Proof

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kuing

1008 posts

kuing

#13 h Dec 1, 2009, 7:00 AM • 1 Y

Y by Adventure10

aakansh92 wrote:

I noticed the following generalisation :
$\sum a^{2} + k \sum ab \geq (k + 1) \sum a \ \forall\ k \in R$

Proof

edit, it seems not true.

This post has been edited 1 time. Last edited by kuing, Aug 28, 2015, 6:38 PM
Reason: Review of old posts, found this proof seems not true.

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Rose_joker

163 posts

Rose_joker

#14 h Dec 3, 2009, 2:15 PM • 1 Y

Y by Adventure10

Let $x = a^3 y = b^3 z = c^3$ the problem is equivalent to
$2\sum_{cyc} a^6 + 18\sum_{cyc} a^3b^3\geq 20\sum_{cyc} a^4bc$
which can be rewritten as
$2\sum_{cyc} a^6 - 2\sum_{cyc} a^3b^3 + 20\sum_{cyc} a^3b^3 - 10\sum_{sym} a^3b^2c\geq 20\sum_{cyc} a^4bc - 10\sum_{sym} a^3b^2c$
or
$\sum_{cyc}((a^2 + ab + b^2)^2 + 10c^3(a + b) - 10abc(a + b))(a - b)^2 \geq 0$
But we have $(a^2 + ab + b^2) + 10c^3(a + b) - 10abc(a + b)\geq 0$ from AM-GM
** $\frac {9}{8}(4a^2b^2 + ab(a + b)^2 + 8c^3(a + b))\geq \frac{27\sqrt [3]{32}}{8}abc(a + b)\geq 10abc(a + b)$
and we have $(a^2 + ab + b^2)^2\geq \frac {9(4a^2b^2 + ab(a + b)^2)}{8}$
Thus, the problem has been solved

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keyree10

249 posts

keyree10

#15 h Feb 1, 2010, 7:41 PM • 2 Y

Y by Adventure10, Mango247

aakansh92 wrote:

If first one occurs we are done and if the second one occurs then replace $a$ by $\frac{1}{a}$ ; $b$ by $\frac{1}{b}$ ; $c$ by $\frac{1}{c}$ and we get the same expression as the first one and thus are done again.

How does this complete the proof ? Doesn't this just show that given a triple $(a,b,c)$ , the inequality holds either for $(a,b,c)$ or for $\left(\frac 1 a,\frac 1 b,\frac 1 c\right)$ ?
Someone please explain this to me.

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