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Inequality with product equal to 1

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Source: Moldova TST Problem 6

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27 posts

Tsarik

#1 h Apr 1, 2015, 5:40 PM • 4 Y

Y by sandumarin, quangminhltv99, Problem_Penetrator, Adventure10

Let $a,b,c$ be positive real numbers such that $abc=1$ . Prove the following inequality:
$a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{9}{2}$ .

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868 posts

#2 h Apr 1, 2015, 7:02 PM • 3 Y

Y by Vladimir_Djurica, Adventure10, Lntkhiem3

Use this lemma
$a^3+b^3+c^3 \ge sum \frac{c(a^2+b^2)}{2}$

and $\frac{ab}{a^2+b^2}=\frac{1}{c(a^2+b^2)}$

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Dr Sonnhard Graubner

16100 posts

Dr Sonnhard Graubner

#3 h Apr 1, 2015, 7:08 PM • 2 Y

Y by Adventure10, Mango247

hello, it can also be proved by BW.
Sonnhard.

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30263 posts

arqady

#4 h Apr 2, 2015, 4:10 AM • 2 Y

Y by Adventure10, Mango247

Dear rightways. Show us a full proof. Thank you!

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641 posts

BSJL

#5 h Apr 2, 2015, 5:39 AM • 3 Y

Y by Problem_Penetrator, Adventure10, Mango247

Hope that I didn't make any stupid mistake.

This problem is equivalent to the following inequality

$\frac{2\left(a^3+b^3+c^3\right)}{abc}+\frac{2ab}{a^2+b^2}+\frac{2bc}{b^2+c^2}+\frac{2ca}{c^2+a^2} \ge 9$

$\Leftrightarrow \frac{(a+b+c)(\left(a-b)^2+(b-c)^2+(c-a)^2\right)}{abc} \ge \frac{(a-b)^2}{a^2+b^2}+\frac{(b-c)^2}{b^2+c^2}+\frac{(c-a)^2}{c^2+a^2}$

$\Leftrightarrow \sum\limits_{cyc}{(a-b)^2\left(\frac{a+b+c}{abc}-\frac{1}{a^2+b^2}\right)}$

Then notice that we have

$\frac{a+b+c}{abc} \ge \frac{1}{ab} \ge \frac{2}{a^2+b^2}$

In conclusion, $\sum\limits_{cyc}{(a-b)^2[\frac{a+b+c}{abc}-\frac{1}{a^2+b^2}]} \ge 0$ , as desired.

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2032 posts

#6 h Apr 2, 2015, 5:55 AM • 2 Y

Y by Adventure10, Mango247

For #1
We have
$\frac{a^3+b^3+c^3}{abc}+\sum{\frac{ab}{a^2+b^2}}-\frac{9}{2}=\sum{\frac{(2a+2b+c)(a-b)^2}{4abc}}+\sum{\frac{(a-b)^4}{4ab(a^2+b^2)}}\ge{0}$

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30263 posts

arqady

#7 h Apr 2, 2015, 7:52 AM • 2 Y

Y by Adventure10, Mango247

The following inequality is stronger.
For positives $a$ , $b$ and $c$ prove that:
$\frac{1}{4}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\geq3$

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641 posts

BSJL

#8 h Apr 2, 2015, 8:04 AM • 3 Y

Y by Problem_Penetrator, Adventure10, Mango247

By using SOS again, we have the following result.

$\frac{1}{4}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\geq3$

$\Leftrightarrow \sum\limits_{cyc}{(a-b)^2\left(\frac{1}{2ab}-\frac{1}{a^2+b^2}\right)} \ge 0$

, which is obvious by AM-GM inequality.

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IstekOlympiadTeam

542 posts

IstekOlympiadTeam

#9 h Apr 2, 2015, 10:49 AM • 4 Y

Y by boss_1998, Problem_Penetrator, tenplusten, Adventure10

Our Solution Anar And Emil Istek Lyceum

Attachments:

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2032 posts

#10 h Apr 2, 2015, 11:23 AM • 2 Y

Y by Adventure10, Mango247

For #7
We have
$\frac{1}{4}\sum{\frac{a+b}{c}}+\sum{\frac{ab}{a^2+b^2}}-3=\sum{\frac{(a-b)^4}{4ab(a^2+b^2)}}\ge{0}$

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166 posts

#11 h Apr 5, 2015, 9:40 AM • 2 Y

Y by Adventure10, Mango247

My Solution:
Notice that $\dfrac{ab}{a^2+b^2}=\dfrac{1}{c(a^2+b^2)}$ since $abc=1$
So we have to prove that
$a^3+b^3+c^3+\frac{1}{c(a^2+b^2)}+\frac{1}{a(b^2+c^2)}+\frac{1}{b(c^2+a^2)} \geq \frac{9}{2}$
Or
$a^3+b^3+c^3+3abc+\frac{1}{c(a^2+b^2)}+\frac{1}{a(b^2+c^2)}+\frac{1}{b(c^2+a^2)} \geq \frac{9}{2}+3$
Applying Schur's inequality
We only have to prove
$\sum{c(a^2+b^2)}+\frac{1}{c(a^2+b^2)}+\frac{1}{a(b^2+c^2)}+\frac{1}{b(c^2+a^2)} \geq \frac{9}{2}+3$
By AM-GM we have
$\dfrac{1}{4}.c(a^2+b^2)+\frac{1}{c(a^2+b^2)}\geq 1$
And since $abc=1$ we have $\sum{c(a^2+b^2)}\geq 6$
Summing all inequalities above, we are done

This post has been edited 1 time. Last edited by tahanguyen98, Apr 5, 2015, 9:43 AM
Reason: Typo fixed!

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130 posts

ISI4

#12 h Apr 5, 2015, 10:27 AM • 2 Y

Y by Adventure10, Mango247

Apply A.M. G.M Inequality twice in the (a^3+b^3+c^3) portion and the other portion .after that In the 2nd portion use rearrangement, then we are done

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215 posts

Jul

#13 h May 3, 2015, 2:49 PM • 1 Y

Y by Adventure10

Tsarik wrote:

Let $a,b,c$ be positive real numbers such that $abc=1$ . Prove the following inequality:
$a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{9}{2}$ .

My solution :
We easily see that :
$\dfrac{ab}{a^2+b^2}=\dfrac{(a+b)^2}{2(a^2+b^2)}-\dfrac{1}{2}$
As a result,
$\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}=\dfrac{(a+b)^2}{2(a^2+b^2)}+\dfrac{(b+c)^2}{2(b^2+c^2)}+\dfrac{(c+a)^2}{2(c^2+a^2)}-\dfrac{3}{2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}-\dfrac{3}{2}$
On the other hand,
$a^3+b^3+c^3\geq \dfrac{(a^2+b^2+c^2)^2}{a+b+c}$
Therefore, we just prove :
$\dfrac{(a^2+b^2+c^2)^2}{a+b+c}+\dfrac{(a+b+c)^2}{a^2+b^2+c^2}\geq 6$
Which is true because of applying AM-GM,
$\dfrac{(a^2+b^2+c^2)^2}{a+b+c}+\dfrac{(a+b+c)^2}{a^2+b^2+c^2}\geq2\sqrt{(a^2+b^2+c^2)(a+b+c)}\geq 2\sqrt{9abc}=6$

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1093 posts

#14 h May 6, 2015, 1:26 PM • 3 Y

Y by blackwave, Adventure10, Mango247

The problem return to:
$a,b,c>0,abc=1\Rightarrow a^3+b^3+c^3-3abc\ge\sum{\frac{1}{2}-\frac{bc}{b^2+c^2}}$
$\Leftrightarrow (b-c)^2[a+b+c-\frac{1}{b^2+c^2}]+(c-a)^2[a+b+c-\frac{1}{c^2+a^2}]+(a-b)^2[a+b+c-\frac{1}{a^2+b^2}]$
Because
$a\ge\frac{1}{b^2+c^2}\Leftrightarrow \frac{1}{bc}\ge\frac{1}{b^2+c^2}\Leftrightarrow bc\le b^2+c^2$
and the other two similar, result it follows that the last inequality is true!
_________
Marin Sandu

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2238 posts

#15 h May 6, 2015, 2:07 PM • 2 Y

Y by Adventure10, Mango247

I am not sure if this is correct though... I'll be leaving out some details

It's easy to show that $a^3+b^3 \geq \frac{(a+b)(a^2+b^2)}{2}$ . (Muirhead's) Thus $a^3+b^3+\frac{2ab}{a^2+b^2} \geq \frac{(a+b)(a^2+b^2)}{2}+\frac{2ab}{a^2+b^2}$ $=2\frac{(a+b)(a^2+b^2)}{4}+\frac{2ab}{a^2+b^2}$ . But by AM-GM this is at least $\frac{^3\sqrt{(a+b)^2(a^2+b^2)ab}}{2}$ .

Again it is easy to show (again by Muirhead's) that this is at least $3ab$ . But then we can develop similar expressions corresponding to $(b,c)$ and $(c,a)$ . We then add up everything; the LHS of the sum is twice the original LHS. Then we get that the LHS of the original ineq is at least $\frac{3(ab+ac+bc)}{2}$ , which is clearly at least $\frac{9}{2}$ by Muirhead's

This post has been edited 1 time. Last edited by Konigsberg, May 6, 2015, 2:08 PM

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801 posts

Sardor

#16 h May 6, 2015, 3:24 PM • 1 Y

Y by Adventure10

We have also $a^3+b^3+c^3 \ge a^2+b^2+c^2$ and by am-gm for six numbers , so we are done!

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42527 posts

sqing

#18 h May 7, 2015, 5:11 AM • 2 Y

Y by Adventure10, Mango247

$a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{1}{2}(a^2+b^2+c^2)+ \frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} +\frac{3}{2}\geq\frac{9}{2}$ .
here

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42527 posts

sqing

#19 h May 9, 2015, 10:06 AM • 3 Y

Y by VadimM, Adventure10, Mango247

Let $t=a^3+b^3+c^3$ , hence $t\geq3\implies (t-3)(2t-3)\geq0\iff t+\frac{9}{2t}\geq \frac{9}{2},$

$a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} =t+\frac{1}{c(a^2+b^2)}+\frac{1}{a(b^2+c^2)}+\frac{1}{b(c^2+a^2)}$

$\geq t+\frac{9}{c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)}\geq t+\frac{9}{2t}\geq \frac{9}{2}.$

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42527 posts

sqing

#20 h May 10, 2015, 7:27 AM • 3 Y

Y by Problem_Penetrator, Adventure10, Mango247

$a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}$
$=a^3+b^3+c^3+\frac{1}{c(a^2+b^2)}+\frac{1}{a(b^2+c^2)}+\frac{1}{b(c^2+a^2)}$
$\geq a^3+b^3+c^3+\frac{9}{c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)}$
$\geq \frac{a^3+b^3+c^3}{2}+\frac{a^3+b^3+c^3}{2}+\frac{9}{2(a^3+b^3+c^3)}$
$\geq 3\sqrt[3]{\frac{a^3+b^3+c^3}{2} \cdot \frac{a^3+b^3+c^3}{2}\cdot \frac{9}{2(a^3+b^3+c^3)}}$
$= \frac{3}{2}\sqrt[3]{9(a^3+b^3+c^3)} \geq \frac{9}{2}.$

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323 posts

adamz

#21 h Jun 16, 2015, 11:40 PM • 2 Y

Y by Adventure10, Mango247

Works?
Click to reveal hidden text

This post has been edited 1 time. Last edited by adamz, Jun 16, 2015, 11:41 PM
Reason: equality

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42527 posts

sqing

#22 h Jul 24, 2015, 3:56 AM • 1 Y

Y by Adventure10

Let $a,b,c$ be positive real numbers such that $abc=1$ . Prove the following inequality $a^3+b^3+c^3+\frac{4ab}{a^2+b^2}+\frac{4bc}{b^2+c^2}+\frac{4ca}{c^2+a^2} \geq 9.$

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VadimM

#23 h Jul 24, 2015, 1:03 PM • 1 Y

Y by Adventure10

Let $a^3+b^3+c^3=t$ hence $t\geq 3$

$ineq= t+\frac{4}{c(a^2+b^2)}+\frac{4}{a(b^2+c^2)}+\frac{4}{b(c^2+a^2)}\geq t+\frac{36}{a^2b+b^2a+c^2b+b^2c+c^2a+a^2c}\geq t+\frac{36}{t+3}\geq 9$
$\iff (t-3)^2\geq 0$

This post has been edited 2 times. Last edited by Kunihiko_Chikaya, Jul 26, 2015, 6:35 PM

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#24 h Sep 11, 2015, 2:41 PM • 2 Y

Y by Adventure10, Mango247

oh, this is my problem!

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Problem_Penetrator

315 posts

Problem_Penetrator

#25 h Mar 16, 2016, 7:24 PM • 3 Y

Y by Mobashereh, Adventure10, Mango247

rightways wrote:

Use this lemma
$a^3+b^3+c^3 \ge sum \frac{c(a^2+b^2)}{2}$

and $\frac{ab}{a^2+b^2}=\frac{1}{c(a^2+b^2)}$

Rightways, using your idea we have:
It's obviously true that $a^{3}+b^{3}+c^{3}\geqslant\sum c(a^{2}+b^{2})$ and that
$\sum\frac{ab}{a^{2}+b^{2}}=\sum\frac{1}{c(a^{2}+b^{2})}$ , so we have left to prove that
$\frac{\sum a(b^{2}+c^{2})}{2}+\sum\frac{1}{c(a^{2}+b^{2})}\geqslant\frac{9}{2}$ . From Cauchy inequality we have:
$\sum\frac{1}{c(a^{2}+b^{2})}\geqslant\frac{9}{\sum c(a^{2}+b^{2})}$ . Now from AM-GM we have that:
$\frac{\sum_{sym}a^{2}b}{2} +\frac{9}{\sum_{sym}a^{2}b}\geqslant 3\sqrt{2}$
This means that your method is too weak for this inequality

This post has been edited 4 times. Last edited by Problem_Penetrator, Mar 16, 2016, 7:56 PM
Reason: Now there are no mystakes

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#26 h Aug 22, 2016, 5:47 PM • 2 Y

Y by Adventure10, Mango247

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#27 h Aug 22, 2016, 7:03 PM • 1 Y

Y by Adventure10

Easy one!
After opening the parentheses and homogenizating (using the fact that $abc=1$ ) we have to prove the following inequality:
$2(\sum_{cyc}a^4b^4c+\sum_{sym}a^4b^3c^2+\sum_{sym}a^7b^2+\sum_{sym}a^5b^4+\sum_{sym}a^3b^4c^2+2\sum_{cyc}a^5b^2c^2) \geq 9(\sum_{sym}a^5b^3c+2a^3b^3c^3)$ .
Using AM-GM we have that:
$\sum_{sym}a^3b^4c^2 \geq 6a^3b^3c^3$
$\sum_{sym}a^5b^4+2\sum_{cyc}a^5b^2c^2 \geq 2(\sum_{sym}a^5b^3c)$
Now our inequality resumes to:
$2(\sum_{cyc}a^4b^4c+\sum_{sym}a^4b^3c^2+\sum_{sym}a^7b^2) \geq 6(\sum_{sym}a^5b^3c) \geq 5(\sum_{sym}a^5b^3c)+6a^3b^3c^3$ .

This post has been edited 2 times. Last edited by rmtf1111, Mar 29, 2018, 2:23 PM

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42527 posts

sqing

#28 h Aug 28, 2017, 9:49 AM • 2 Y

Y by bel.jad5, Adventure10

https://artofproblemsolving.com/community/c6h1503751p8887849
https://artofproblemsolving.com/community/c6h1503753p8887869:
Let $a$ , $b$ , $c$ be positive real numbers such that $abc=1$ . Prove that
$a^{3}+b^{3}+c^{3}+\frac{ab}{a^{2}+b^{2}}+\frac{bc}{b^{2}+c^{2}}+\frac{ac}{c^{2}+a^{2}}\geq \frac{9}{5}(a^2+b^2+c^2-\frac{1}{2})\geq \frac{13}{2}(a+b+c)-15\geq 2(a+b+c)-\frac{3}{2}\geq \frac{9}{2}$

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Quantum_fluctuations

1282 posts

Quantum_fluctuations

#29 h Aug 26, 2019, 6:00 PM • 3 Y

Y by teomihai, Adventure10, Mango247

Jul wrote:

Tsarik wrote:

Let $a,b,c$ be positive real numbers such that $abc=1$ . Prove the following inequality:
$a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{9}{2}$ .

My solution :
We easily see that :
$\dfrac{ab}{a^2+b^2}=\dfrac{(a+b)^2}{2(a^2+b^2)}-\dfrac{1}{2}$ As a result,
$\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}=\dfrac{(a+b)^2}{2(a^2+b^2)}+\dfrac{(b+c)^2}{2(b^2+c^2)}+\dfrac{(c+a)^2}{2(c^2+a^2)}-\dfrac{3}{2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}-\dfrac{3}{2}$ On the other hand,
$a^3+b^3+c^3\geq \dfrac{(a^2+b^2+c^2)^2}{a+b+c}$ Therefore, we just prove :
$\dfrac{(a^2+b^2+c^2)^2}{a+b+c}+\dfrac{(a+b+c)^2}{a^2+b^2+c^2}\geq 6$ Which is true because of applying AM-GM,
$\dfrac{(a^2+b^2+c^2)^2}{a+b+c}+\dfrac{(a+b+c)^2}{a^2+b^2+c^2}\geq2\sqrt{(a^2+b^2+c^2)(a+b+c)}\geq 2\sqrt{9abc}=6$

ELEGANT!

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3750 posts

#30 h Aug 26, 2019, 7:32 PM • 1 Y

Y by Adventure10

Tsarik wrote:

Let $a,b,c$ be positive real numbers such that $abc=1$ . Prove the following inequality:
$a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{9}{2}$ .

Because:
$LHS \geq \frac{a^3+b^3+c^3}{2}+\sum_{cyc}\left(\frac{a^2+b^2}{4}+\frac{ab}{a^2+b^2}\right)\geq \frac{a^3+b^3+c^3}{2}+\sum_{cyc} \sqrt{ab} \geq \frac{3}{2}+3=\frac{9}{2}=RHS$

This post has been edited 2 times. Last edited by bel.jad5, Aug 26, 2019, 8:06 PM

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sqing

#31 h Apr 29, 2021, 6:54 AM

Y by

Tsarik wrote:

Let $a,b,c$ be positive real numbers such that $abc=1$ . Prove the following inequality:
$a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{9}{2}$ .

T6/444.

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406 posts

#32 h Sep 27, 2022, 12:30 PM

Y by

Tsarik wrote:

Let $a,b,c$ be positive real numbers such that $abc=1$ . Prove the following inequality:
$a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{9}{2}$ .

$LHS+3\geqslant 2\sum a^2+\sum \frac{ab}{a^2+b^2}=\sum \frac{a^2+b^2}{4}+\frac{ab}{a^2+b^2}+\frac{3}{2}\sum a^2\geq \sum \sqrt{ab}+\frac{3}{2}\sum a^2\geqslant 3+\frac{9}{2}=RHS+3$

This post has been edited 1 time. Last edited by P2nisic, Sep 27, 2022, 12:30 PM

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#36 h Apr 16, 2025, 2:25 PM

Y by

$\frac{a^3+b^3}{2}+\frac{ab}{a^2+b^2}\geqslant\frac{3ab}{2}$

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