Cyclic sum a^2b(b-c) / (a+b) >= 0

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sandu2508

152 posts

sandu2508

#1 h May 4, 2010, 12:14 PM • 4 Y

Y by mathjeka, Adventure10, lian_the_noob12, and 1 other user

Let $a,b$ and $c$ be positive real numbers. Prove that $\frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0.$

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hflowers

3 posts

hflowers

#2 h May 4, 2010, 12:31 PM • 4 Y

Y by yugrey, Adventure10, Mango247, and 1 other user

very easy problem . CLear arithmetic, and AM-GM at the end

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wya

257 posts

wya

#3 h May 4, 2010, 12:36 PM • 10 Y

Y by SpectrumOfA, pablock, xyzz, Quidditch, ehuseyinyigit, Adventure10, Mango247, and 3 other users

sandu2508 wrote:

Let $a,b$ and $c$ be positive real numbers. Prove that

$\frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0$

Substitution $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c},$
this inequality become
$\sum\frac{x-z}{y+z}\ge0,$
which is true by Rearrangement Inequality
$\sum x\left(\frac{1}{y+z}\right)\ge\sum z\left(\frac{1}{y+z}\right).$

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turcas_c

201 posts

turcas_c

#4 h May 4, 2010, 2:09 PM • 5 Y

Y by Sx763_, Problem_Penetrator, Adventure10, Mango247, and 1 other user

An ugly solution:

If somebody had the patience to make the common denominator, he would had a surprise:

$\frac{a^{2}b(b-c)}{a+b}+\frac{b^{2}c(c-a)}{b+c}+\frac{c^{2}a(a-b)}{c+a} = \frac{a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2}{(a+b)(b+c)(c+a)}.$

We have to prove that:

$a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2 \geq 0 \Leftrightarrow$

$a^3b^3+b^3c^3+c^3a^3 \geq a^3b^2c+a^2bc^3+ab^3c^2$ .

We can see that this inequality is true by Muirhead, and prove it taking:

$\frac{a^3b^3+a^3b^3+b^3c^3}{3} \geq a^2b^3c$ (by AM-GM).

and its analogs. Summing up those 3 inequalities, our problem is solved.

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Bugi

1857 posts

Bugi

#5 h May 4, 2010, 2:10 PM • 2 Y

Y by Adventure10, Mango247

Wow easy. Rearrangement is so obvious, the differences b-c,c-a,a-b immediately reminded me of it (wya, excellent solution, I never thought of that substitution).

I thought like this: Dividing everything with abc gets rid of ugly $a^2b,b^2c,c^2a$ terms and leaves us only with $ab,bc,ca$ to work with.

So, if some of $a,b,c$ is zero, easy.
Else divide everything with abc, and set substitute $x=bc,y=ca,z=ab$

We get the inequality $\sum \frac{z-y}{x+y}\ge 0$ which is true by Rearrangement (wya's work)

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Anni

74 posts

Anni

#6 h May 4, 2010, 6:15 PM • 3 Y

Y by Problem_Penetrator, Adventure10, Mango247

very easy like this :if we sum up abc to each term of left hand side and also 3abc to the right hand side and then we apply the a.m-g.m...all is done

This post has been edited 1 time. Last edited by Anni, May 5, 2010, 6:44 PM

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marin.bancos

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marin.bancos

#7 h May 4, 2010, 7:01 PM • 6 Y

Y by hehe2, Problem_Penetrator, ehuseyinyigit, Adventure10, Mango247, and 1 other user

Let's do it, dear Anni!
The inequality is equivalent to
$abc+\frac{a^{2}b(b-c)}{a+b}+abc+\frac{b^{2}c(c-a)}{b+c}+abc+\frac{c^{2}a(a-b)}{c+a}\ge 3abc$
Therefore
$\left.LHS=ab^2\cdot\frac{a+c}{a+b}+bc^2\cdot\frac{b+a}{b+c}+ca^2\cdot\frac{c+b}{c+a}\stackrel{AM-GM}{\geq}3\sqrt[3]{ab^2\cdot\frac{a+c}{a+b}\cdot bc^2\cdot\frac{b+a}{b+c}\cdot ca^2\cdot\frac{c+b}{c+a}}=3abc=RHS\right.$

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chinacai

254 posts

chinacai

#8 h May 5, 2010, 12:07 PM • 2 Y

Y by Adventure10, Mango247

What is BMO？ Balank or British

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marin.bancos

725 posts

marin.bancos

#9 h May 5, 2010, 12:12 PM • 2 Y

Y by Adventure10, Mango247

It's Balkan!

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NickNafplio

422 posts

NickNafplio

#10 h May 5, 2010, 3:58 PM • 5 Y

Y by andrejilievski, Problem_Penetrator, kiyoras_2001, Adventure10, Mango247

There are many ways to solve this!

Another one is:

Equivalently we have:

$\sum_{cyclic}{\frac{ab}{bc + ca}} \geq \sum_{cyclic}{\frac{a}{b + a}}$

Add $\sum_{cyclic}{\frac{b}{b + a}}$ to both sides

Then we need to show:

$\sum_{cyclic}{\frac{ab}{ac + bc} + \frac{b}{b + a}} = \sum_{cyclic}{\frac{ab}{ac + bc} + \frac{bc}{bc + ac}} = \sum_{cyclic}{\frac{ab + bc}{ac + bc}} \geq 3$ which is true because by AM-GM:

$\sum_{cyclic}{\frac{ab + bc}{ac + bc}} \geq 3\sqrt[3]{\frac{(ab + bc)(bc + ca)(ca + ab)}{(ab + ca)(ca + bc)(bc + ab)}} = 3$ .

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Anni

74 posts

Anni

#11 h May 6, 2010, 7:20 PM • 2 Y

Y by Adventure10, Mango247

which state proposed this problem?
was any problem in the exam proposed by albanian leader?

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Thebattlefront

184 posts

Thebattlefront

#12 h May 7, 2010, 1:16 AM • 2 Y

Y by Adventure10, Mango247

wya may I ask something?

When you substitute the $x = 1/a$ , $y = 1/b$ , $z = 1/c$ , did you have to write out, by hand, the transformation from the original substitution into this?:

$\sum\frac{x-z}{y+z}\ge0$

Or is there some property of summations that allows you to make easy work of such a substitution? I did it by hand using the first term $\frac{a^{2}b(b - c)}{a + b}$ and got

$\sum\frac{z-y}{y+x}\ge0$

which is equivalent to yours, however, I was wondering why you did not end up with the same looking summation as I had, and thus, was wondering if there was some special way of doing this.

and can someone show me the rearrangement part?

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stolet

3 posts

stolet

#13 h May 7, 2010, 7:08 AM • 2 Y

Y by Adventure10, Mango247

The substitution that wya used is first that try most of the competitors leading us to the equivalent inequality $\sum\frac{x-z}{y+z}\ge0$
If we consider that the sum of the nominators is zero, we can continue even without using a theorem. Then we can re-write in obvious forms:

if x is maximum(or one of the maximums) using $(y-x)=-((z-y)+(x-z))$ : $\frac{(z-y)^{2}}{(x+y)(x+z)}+\frac{(x-z)(x-y)}{(y+z)(x+z)}\ge 0$ and it is clear that inequality holds.
Using same idea, re-writing other two nominators, when y or z is maximum, we can obtain some similar forms.

Thus, it holds in every case.

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hedeng123

485 posts

hedeng123

#14 h May 7, 2010, 8:06 AM • 1 Y

Y by Adventure10

sandu2508 wrote:

Let $a,b$ and $c$ be positive real numbers. Prove that

$\frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0$

$\Leftrightarrow \frac{c^3(2a+b)(a-b)^2+a^3(2b+c)(b-c)^2+b^3(2c+a)(c-a)^2}{3}\ge 0$

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broniran

142 posts

broniran

#15 h May 7, 2010, 8:51 AM • 2 Y

Y by Adventure10, Mango247

turcas_c wrote:

An ugly solution:

If somebody had the patience to make the common denominator, he would had a surprise:

$\frac{a^{2}b(b-c)}{a+b}+\frac{b^{2}c(c-a)}{b+c}+\frac{c^{2}a(a-b)}{c+a} = \frac{a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2}{(a+b)(b+c)(c+a)}.$

We have to prove that:

$a^3b^3-a^3b^2c+a^3c^3-a^2bc^3+b^3c^3-ab^3c^2 \geq 0 \Leftrightarrow$

$a^3b^3+b^3c^3+c^3a^3 \geq a^3b^2c+a^2bc^3+ab^3c^2$ .

We can see that this inequality is true by Muirhead, and prove it taking:

$\frac{a^3b^3+a^3b^3+b^3c^3}{3} \geq a^2b^3c$ (by AM-GM).

and its analogs. Summing up those 3 inequalities, our problem is solved.

I proved it just like you and I believe this is the easiest and most elegant way of proving this inequality.

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orthocentre

72 posts

orthocentre

#49 h Nov 8, 2018, 12:15 PM • 1 Y

Y by Adventure10

With the rearrangement inequality, we know that

$x^3+y^3+z^3\ge x^2y+y^2z+z^2x$
for all positive real $x, y, z$ .

Now we let $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$ so we know that

$\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\ge\frac{1}{bc^2}+\frac{1}{ca^2}+\frac{1}{ab^2}$
for all positive real $a, b, c$ .

Multiplying both sides by $a^3b^3c^3$ we get

$a^3b^3+b^3c^3+c^3a^3\ge a^3b^2c+b^3c^2a+c^3a^2b.$
Therefore $a^3b^3+b^3c^3+c^3a^3-a^3b^2c-b^3c^2a-c^3a^2b\ge0$ .

Adding some stuff and taking it away again, we get

$a^2b^3c-a^2bc^3+a^3b^3-a^3bc^2+ab^2c^3+b^3c^3-a^3b^2c-a^2b^3c+a^3bc^2-ab^3c^2+a^3c^3-ab^2c^3 \ge 0.$
Oh, look! It magically factorises to

$a^2b(b^2-c^2)(c+a)+b^2(c^2-a^2)(a+b)+c^2a(a^2-b^2)(b+c)\ge0.$
I bet you're wondering how I found such a beautiful factorisation. Well, now, you can divide both sides by $(a+b)(b+c)(c+a)$ to get

$\frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0.$
To find how I worked this (pretty boring) method out, just follow all my steps in reverse.

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Feridimo

563 posts

Feridimo

#50 h Jan 11, 2020, 4:33 PM • 2 Y

Y by Adventure10, Mango247

hint

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sqing

41825 posts

sqing

#51 h Apr 7, 2020, 6:45 AM

Y by

sandu2508 wrote:

Let $a,b$ and $c$ be positive real numbers. Prove that $\frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0.$

Let $a,b,c>0$ . Prove that $\frac{a(c-b)}{2a+3b+c}+\frac{b(a-c)}{2b+3c+a}+\frac{c(b-a)}{2c+3a+b}\ge 0$ https://artofproblemsolving.com/community/c6h489134p2741648
$\frac{(a+2b+1)(a-b)}{b(2c+a)}+\frac{(b+2c+1)(b-c)}{c(2a+b)}+\frac{(c+2a+1)(c-a)}{a(2b+c)}\ge0$ https://artofproblemsolving.com/community/c6h576961p3401520
$\frac{a(a-b)}{c^2+ca+ab}+\frac{b(b-c)}{a^2+ab+bc}+\frac{c(c-a)}{b^2+bc+ca}\geq 0$ https://artofproblemsolving.com/community/c6h583859p3451999
For $a, b, c>0$ , prove the following "twin"" inequalities:
$\frac{ab(b^2-ac)}{a+b}+\frac{bc(c^2-ab)}{b+c}+\frac{ca(a^2-bc)}{c+a} \ge 0.$ $\frac{ab(ab-c^2)}{b+c}+\frac{bc(bc-a^2)}{c+a}+\frac{ca(ca-b^2)}{a+b} \ge 0.$ https://artofproblemsolving.com/community/c6h2094743p15141068

This post has been edited 1 time. Last edited by sqing, May 10, 2020, 8:49 AM

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Keith50

464 posts

Keith50

#53 h Mar 17, 2021, 2:16 AM

Y by

sandu2508 wrote:

Let $a,b$ and $c$ be positive real numbers. Prove that $\frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0.$

By clearing the denominators, we have to show that $\sum_{cyc} a^2b(b^2-c^2)(c+a)=\sum_{cyc} a^3b^3-\sum_{cyc} a^3b^2c\ge 0$ which by Schur's Inequality, we know that $\sum_{cyc} a^3b^3 \ge \sum_{sym} a^3b^2c - 3a^2b^2c^2\ge \sum_{cyc} a^3b^2c$ $\textbf{\textit{IFF}}$ $\sum_{cyc} a^2c \ge 3abc$ which is just AM-GM. $\quad \blacksquare$

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Assassino9931

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Assassino9931

#54 h Jan 7, 2022, 2:02 PM • 1 Y

Y by ehuseyinyigit

By adding $abc$ to each fraction, we wish to prove $\sum \frac{ab^2(a+c)}{a+b} \geq 3abc$ which follows directly from AM-GM.

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sqing

41825 posts

sqing

#55 h Jan 7, 2022, 2:38 PM

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Let $a,b$ and $c$ be positive real numbers. Prove that $\frac{a^2b(ab-c^2)}{a+b}+\frac{b^2c(bc-a^2)}{b+c}+\frac{c^2a(ca-b^2)}{c+a} \ge 0$

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arqady

30214 posts

arqady

#56 h Jan 7, 2022, 4:40 PM

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sqing wrote:

Let $a,b$ and $c$ be positive real numbers. Prove that $\frac{a^2b(ab-c^2)}{a+b}+\frac{b^2c(bc-a^2)}{b+c}+\frac{c^2a(ca-b^2)}{c+a} \ge 0$

It's $\sum_{cyc}(a^4b^3-a^3b^2c^2)\geq0,$ which is obvious.

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sqing

41825 posts

sqing

#57 h Jan 8, 2022, 12:01 AM

Y by

Thanks.

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Mirhabib

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Mirhabib

#58 h Feb 12, 2023, 7:24 AM

Y by

sandu2508 wrote:

Let $a,b$ and $c$ be positive real numbers. Prove that $\frac{a^2b(b-c)}{a+b}+\frac{b^2c(c-a)}{b+c}+\frac{c^2a(a-b)}{c+a} \ge 0.$

if we divide each side by abc, we get
$\sum(\frac{ab-ac}{ac+bc})\ge 0$
$ab=x, bc=y, ac=z$
$\sum(\frac{x-z}{y+z})\ge 0$
add 3 both sides,
$1-\frac{z}{y+z}=\frac{y}{y+z}$ so, $\sum(\frac{x+y}{y+z})\ge 3$ and it is well-known(according to AM-GM). So, we are done.

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a_0a

34 posts

a_0a

#59 h Oct 28, 2023, 5:35 PM

Y by

Solution

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ehuseyinyigit

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ehuseyinyigit

#60 h Nov 13, 2023, 11:46 PM

Y by

Generalization 1
Let $a_{1},a_{2},\cdots,a_{n}$ be positive reals ( $n\geq 3$ ). Then prove that

$\sum_{cyc}{\left(\dfrac{\prod{\left(a_{1}\right)}\dfrac{a_{k}}{a_{k-1}}\left(a_{k+1}-a_{k-1}\right)}{a_{k}+a_{k+1}}\right)}\geq 0$

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Draq

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Draq

#61 h Nov 11, 2024, 6:44 PM • 1 Y

Y by MuradSafarli

First sum it and after that it can be done easily by muirhead ineq

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BreezeCrowd

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BreezeCrowd

#62 h Nov 17, 2024, 7:42 PM

Y by

MexicOMM wrote:

Sorry to revive, but I think this solution is the simplest one. Like everybody has seen, we need to prove that:
$\frac{a^2b^2}{a+b} + \frac{a^2c^2}{a+c} + \frac{b^2c^2}{b+c} \ge \frac{a^2bc}{a+b} +\frac{abc^2}{a+c}+\frac{ab^2c}{b+c}$ WLOG we can asume that $a \ge b \ge c$ , so we get $\frac{ab}{a+b} \ge \frac{ac}{a+c} \ge \frac{cb}{c+b}$ , you can see this by just simplifying, and $ab \ge ac \ge bc$ , so applying the rearrangement inequality, with the previous sets we get the result.

Why we can take $a\ge b\ge c$ ?
This inequality not symmetric

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arqady

30214 posts

arqady

#63 h Nov 17, 2024, 8:36 PM

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BreezeCrowd wrote:

MexicOMM wrote:

Sorry to revive, but I think this solution is the simplest one. Like everybody has seen, we need to prove that:
$\frac{a^2b^2}{a+b} + \frac{a^2c^2}{a+c} + \frac{b^2c^2}{b+c} \ge \frac{a^2bc}{a+b} +\frac{abc^2}{a+c}+\frac{ab^2c}{b+c}$ WLOG we can asume that $a \ge b \ge c$ , so we get $\frac{ab}{a+b} \ge \frac{ac}{a+c} \ge \frac{cb}{c+b}$ , you can see this by just simplifying, and $ab \ge ac \ge bc$ , so applying the rearrangement inequality, with the previous sets we get the result.

Why we can take $a\ge b\ge c$ ?
This inequality not symmetric

It means triples $(ab,ac,bc)$ and $\left(\frac{ab}{a+b},\frac{ac}{a+c},\frac{cb}{c+b}\right)$ have the same ordering.

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AshAuktober

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AshAuktober

#64 h Apr 8, 2025, 1:13 PM • 1 Y

Y by khan.academy

Expanding this (and its not hard) we get the ineq to be equivalent to $\sum_{cyc} a^3b^3 \ge \sum_{cyc} a^2bc^3.$ But from Weighted AM-GM, $\frac{b^3a^3+2b^3c^3}{3} \ge b^3ac^2,$ and adding gives the desired result. $\square$

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