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Source: Zhautykov Olympiad, Kazakhstan 2008, Question 6

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207 posts

#1 h Jan 18, 2008, 4:49 PM • 2 Y

Y by Adventure10, Mango247

Let $a, b, c$ be positive integers for which $abc = 1$ . Prove that
$\sum \frac{1}{b(a+b)} \ge \frac{3}{2}$ .

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1394 posts

nayel

#2 h Jan 18, 2008, 5:28 PM • 3 Y

Y by vsathiam, Adventure10, Mango247

I think you mean positive reals.

This is very easy, I can't believe it was proposed in a recent olympiad..

Letting $a = \frac xy,b = \frac yz, c = \frac zx$ implies
$LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x}$
Now it remains to prove that
$2(x^2 + y^2 + z^2)^2\ge 3\sum_{cyc}x^2y^2 + 3\sum_{cyc}x^3y$
Which follows by adding the two inequalities
$x^4 + y^4 + z^4\ge x^3y + y^3z + z^3x \\ \\ \sum_{cyc}(x^4 + x^2y^2)\ge \sum_{cyc}2x^3y$
I bet all the contestants solved this one.

EDIT: Here is a related topic: http://www.mathlinks.ro/viewtopic.php?search_id=1111061624&t=165948

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3183 posts

#3 h Jan 19, 2008, 3:53 AM • 1 Y

Y by Adventure10

the reason u r getting the solution easily is u have taken the LHS wrong...
it should be $\sum \frac {xz}{x^{2} + yz}$

P.S. laso how did u get ur first inequality that is

nayel wrote:

$LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x}$

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1363 posts

Erken

#4 h Jan 19, 2008, 6:40 AM • 3 Y

Y by Crazy4Hitman, Adventure10, Mango247

pardesi wrote:

the reason u r getting the solution easily is u have taken the LHS wrong...
it should be $\sum \frac {xz}{x^{2} + yz}$

No,nayel's solution is right.

nayel wrote:

$LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x}$

He used Cauchy-Schwarz ineqaulity

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3183 posts

#5 h Jan 19, 2008, 6:44 AM • 2 Y

Y by Adventure10, Mango247

sorry i had copied the inequality wrong

sorry nayel

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3183 posts

#6 h Jan 19, 2008, 6:50 AM • 2 Y

Y by Adventure10, Mango247

and how is that from cauchy

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1363 posts

Erken

#7 h Jan 19, 2008, 6:53 AM • 2 Y

Y by Adventure10, Mango247

pardesi wrote:

and how is that from cauchy

$\sum\frac{x^2}{z^2+xy}=\sum\frac{x^4}{z^2x^2+x^3y}=\sum\frac{(x^2)^2}{z^2x^2+x^3y}$ .
P.S:
Official solution is much nicer and easier!
I hope that some of you will find it.

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3183 posts

#8 h Jan 19, 2008, 7:07 AM • 2 Y

Y by Adventure10, Mango247

isn't it strange if we apply the engel form to
$\sum \frac{x^{2}}{z^{2}+yx} \geq \frac{(\sum x)^{2}}{\sum x^{2} + \sum xy}$
we don't get the desired result...we get a weak bound.

u have to multiply by $x^{2}$

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1363 posts

Erken

#9 h Jan 19, 2008, 3:53 PM • 2 Y

Y by Adventure10, Mango247

pardesi wrote:

isn't it strange if we apply the engel form to
$\sum \frac {x^{2}}{z^{2} + yx} \geq \frac {(\sum x)^{2}}{\sum x^{2} + \sum xy}$
we don't get the desired result...we get a weak bound.

u have to multiply by $x^{2}$

Yes,that is even mystical.

That is the reason why,i didn't get 7 points for this problem,
but anyway i've received gold medal

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3183 posts

#10 h Jan 19, 2008, 4:11 PM • 2 Y

Y by Adventure10, Mango247

congrats anyways

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1394 posts

nayel

#11 h Jan 19, 2008, 5:18 PM • 2 Y

Y by Adventure10, Mango247

pardesi wrote:

isn't it strange if we apply the engel form to
$\sum \frac {x^{2}}{z^{2} + yx} \geq \frac {(\sum x)^{2}}{\sum x^{2} + \sum xy}$
we don't get the desired result...we get a weak bound.

u have to multiply by $x^{2}$

Multiplying by $x^2$ isn't necessary. It's enough to multiply by $x$ . (But $x^2$ makes things a bit more easier)
$\sum_{cyc}\frac {x^3}{z^2x + x^2y}\ge\frac {(x^{\frac 32} + y^{\frac 32} + z^{\frac 32})^2}{2(x^2y + y^2z + z^2x)}$
Then it remains to show that
$(x^{\frac 32} + y^{\frac 32} + z^{\frac 32})^2\ge 3(x^2y + y^2z + z^2x)$
Which follows from AM-GM
$\sum_{cyc}(x^3 + x^{\frac 32}y^{\frac 32} + x^{\frac 32}y^{\frac 32})\ge \sum_{cyc}3x^2y$
PS. Congrats to Erken.

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3183 posts

#12 h Jan 19, 2008, 5:38 PM • 2 Y

Y by Adventure10, Mango247

no it's not multiplying by $x$ or $x^{2}$ ...the thing is u can't put engel form directly without multiplying and dividing out the same...and get the answer

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1394 posts

nayel

#13 h Jan 19, 2008, 5:53 PM • 2 Y

Y by Adventure10, Mango247

pardesi wrote:

...the thing is u can't put engel form directly without multiplying and dividing out the same...and get the answer

That's what makes the problem a bit more standard. Otherwise it would be trivial right?

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3183 posts

#14 h Jan 19, 2008, 5:55 PM • 2 Y

Y by Adventure10, Mango247

see my point was not about triviality...but the fact that until and unless u multiply it by $x$ or $x^{2}$ or so u don't get the solution is n't that amazing

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1394 posts

nayel

#15 h Jan 19, 2008, 5:59 PM • 2 Y

Y by Adventure10, Mango247

Yes you're right that is interesting... but it's also a well known strategy.

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204 posts

lasha

#16 h Jan 25, 2008, 2:34 PM • 3 Y

Y by Evenprime123, Adventure10, Mango247

OFFICIAL SOLUTION:
LEMMA: Suppose that $x_{1} > = x_{2} > = x_{3},y_{1} < = y_{2} < = y_{3}$ , and $(y_{1}^{'};y_{2}^{'};y_{3}^{'})$ is a permutation of $(y_{1};y_{2};y_{3})$ .Then $x_{1}y_{1}^{'} + x_{2}y_{2}^{'} + x_{3}y_{3}^{'} > = x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3}$ .
Proof: Since $y_{1} < = y_{2} < = y_{3}$ , we get $y_{1}^{'} > = y_{1}$ and $y_{1}^{'} + y_{2}^{'} > = y_{1} + y_{2}$ ;moreover, $y_{1}^{'} + y_{2}^{'} + y_{3}^{'} = y_{1} + y_{2} + y_{3}$ .
$x_{1}y_{1}^{'} + x_{2}y_{2}^{'} + x_{3}y_{3}^{'} = (x_{1} - x_{2})y_{1}^{'} + (x_{2} - x_{3})(y_{1}^{'} + y_{2}^{'}) + x_{3}(y_{1}^{'} + y_{2}^{'} + y_{3}^{'}) > = (x_{1} - x_{2})y_{1} + (x_{2} - x_{3})(y_{1} + y_{2}) + x_{3}(y_{1} + y_{2} + y_{3}) = x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3}$ , Q.E.D.
Denote by $S$ the left-hand part of the desired inequality.Since $S$ is invariant under the cyclical permutation of variables, we can assume $a < = b < = c$ or $a > = b > = c$ . In both cases, by LEMMA we get: $S = \frac {1}{(a + b)b} + \frac {1}{(b + c)c} + \frac {1}{(c + a)a} > = \frac {1}{(a + b)c} + \frac {1}{(b + c)a} + \frac {1}{(c + a)b} = T$ . Hence,
$2S > = S + T = (\frac {1}{(a + b)c} + \frac {1}{(a + b)b}) + (\frac {1}{(b + c)a} + \frac {1}{(b + c)c}) + (\frac {1}{(c + a)b} + \frac {1}{(c + a)a}) = \frac {b + c}{(a + b)bc} + \frac {c + a}{(b + c)ca} + \frac {a + b}{(c + a)ab} > = 3(\frac {1}{abc})^{\frac {1}{3}} = 3$ , Q.E.D.

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Umut Varolgunes

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Umut Varolgunes

#17 h Jan 26, 2008, 9:14 AM • 2 Y

Y by Adventure10, Mango247

$\sum\frac{x^2}{y^2+xz}\geq\sum\frac{2x^2}{2y^2+x^2+z^2}$ from A.G.M and now applying C.S. in Engel form works
$\sum\frac{2x^2}{2y^2+x^2+z^2}\geq\frac{2(x^2+y^2+z^2)^2}{x^4+y^4+z^4+3x^2y^2+3y^2z^2+3x^2z^2}\geq\frac{2.3}{4}$

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2861 posts

Vasc

#18 h Jan 26, 2008, 2:00 PM • 2 Y

Y by Adventure10, Mango247

This is harder.

Let $a, b, c$ be positive numbers for which $abc = 1$ . Prove that
$\sum \frac {1}{b(5a +b)} \ge \frac {1}{2}$ .

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253 posts

ehku

#19 h Jan 26, 2008, 2:23 PM • 2 Y

Y by Adventure10, Mango247

Vasc wrote:

This is harder.

Let $a, b, c$ be positive numbers for which $abc = 1$ . Prove that
$\sum \frac {1}{b(5a + b)} \ge \frac {1}{2}$ .

Put $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ where $x,y,z >0$ , we have to prove
$\sum \frac{z^2}{y^2+5xz} \ge \frac {1}{2}$
By Cauchy-Schwarz inequality:
$\sum \frac{z^2}{y^2+5xz}= \sum \frac{z^4}{y^2z^2+5xz^3} \ge \frac{(x^2+y^2+z^2)^2}{\sum x^2y^2 +5 \sum x^3y}$
It suffits to prove
${ 2(x^2+y^2+z^2)^2 \ge \sum x^2y^2 +5 \sum x^3y}$ ,
which is true by your inequality $(x^2+y^2+z^2)^2 \ge 3(x^3y+y^3z+z^3x)$ and the inequality $(x^2+y^2+z^2)^2 \ge 3(x^2y^2+y^2z^2+z^2x^2)$ .

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nayel

#20 h Jan 26, 2008, 5:55 PM • 1 Y

Y by Adventure10

Actually if anyone saw my link above it shows that the following one holds for all $a,b,c,m,n$ satisfying $abc = 1$
$\sum_{cyc}\frac {1}{b(ma + nb)}\ge\frac {3}{m + n}$
And it's easily proven using only Cauchy-Schwarz inequality.

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560 posts

#21 h Dec 20, 2011, 4:37 AM • 2 Y

Y by Adventure10, Mango247

EAsy TRY REARRANGEMENT ineq

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312 posts

#22 h Nov 18, 2015, 8:26 PM • 3 Y

Y by samirka259, Adventure10, Mango247

PhilAndrew wrote:

Let $a, b, c$ be positive integers for which $abc = 1$ . Prove that
$\sum \frac{1}{b(a+b)} \ge \frac{3}{2}$ .

Let's write the $LHS$ as $\frac{c^2a}{c^2ab(a+b)}+\frac{b^2c}{b^2ca(c+a)}+\frac{a^2b}{a^2bc(b+c)}$ , equivalently $\frac{c^2a}{c(a+b)}+\frac{b^2c}{b(c+a)}+\frac{a^2b}{a(b+c)}$ (since $abc=1$ ). Then using Cauchy-Schwarz inequality we have $\frac{c^2a}{c(a+b)}+\frac{b^2c}{b(c+a)}+\frac{a^2b}{a(b+c)}\geq \frac{(c\sqrt{a}+b\sqrt{c}+a\sqrt{b})^2}{2(ab+bc+ca)}.$ So, we have to show that $(c\sqrt{a}+b\sqrt{c}+a\sqrt{b})^2\geq 3(ab+bc+ca).$ Expanding and using AM-GM inequality, $abc=1$ we get
$(c\sqrt{a}+b\sqrt{c}+a\sqrt{b})^2=c^2a+b^2c+a^2b+2(bc\sqrt{ca}+ca\sqrt{ab}+ab\sqrt{bc})=c^2a+b^2c+a^2b+2(c\sqrt{b}+a\sqrt{c}+b\sqrt{a})=(c^2a+a\sqrt{c}+a\sqrt{c})+(b^2c+c\sqrt{b}+c\sqrt{b})+(a^2b+b\sqrt{a}+b\sqrt{a})\geq 3ca+3bc+3ab=3(ab+bc+ca).$ So, the inequality is proven.

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2032 posts

#23 h Nov 19, 2015, 12:29 AM • 3 Y

Y by Adventure10, Mango247, MELSSATIMOV40

PhilAndrew wrote:

Let $a, b, c$ be positive integers for which $abc = 1$ . Prove that
$\sum \frac{1}{b(a+b)} \ge \frac{3}{2}$ .

We have
$LHS-RHS=\frac{1}{6(a+b)(b+c)(c+a)}\sum{\frac{(6ab^2c+abc+2bc+3a)(a-1)^2}{ca}}+\frac{1}{2(a+b)(b+c)(c+a)}\sum{\frac{(a-1)^2(b-1)^2}{a}}\ge{0}$

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30146 posts

arqady

#24 h Apr 29, 2016, 8:37 AM • 2 Y

Y by Adventure10, Mango247

PhilAndrew wrote:

Let $a, b, c$ be positive integers for which $abc = 1$ . Prove that
$\sum \frac{1}{b(a+b)} \ge \frac{3}{2}$ .

It was here:
http://artofproblemsolving.com/community/c6h81588

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#26 h Sep 3, 2017, 8:54 AM • 1 Y

Y by Adventure10

nayel wrote:

I think you mean positive reals.

This is very easy, I can't believe it was proposed in a recent olympiad..

Letting $a = \frac xy,b = \frac yz, c = \frac zx$ implies
$LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x}$ Now it remains to prove that
$2(x^2 + y^2 + z^2)^2\ge 3\sum_{cyc}x^2y^2 + 3\sum_{cyc}x^3y$ Which follows by adding the two inequalities
$x^4 + y^4 + z^4\ge x^3y + y^3z + z^3x \\ \\ \sum_{cyc}(x^4 + x^2y^2)\ge \sum_{cyc}2x^3y$ I bet all the contestants solved this one.

EDIT: Here is a related topic: http://www.mathlinks.ro/viewtopic.php?search_id=1111061624&t=165948

i think the last inequality x^4 + y^4 + z^4 >= x^3y +y^3z +z^3x.... (it's true according to rearrangement) but in this question a.b.c = 1
so i think we can't work

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769 posts

#28 h Apr 7, 2018, 1:59 AM • 2 Y

Y by Adventure10, Mango247

ehku wrote:

Vasc wrote:

This is harder.

Let $a, b, c$ be positive numbers for which $abc = 1$ . Prove that
$\sum \frac {1}{b(5a + b)} \ge \frac {1}{2}$ .

Put $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ where $x,y,z >0$ , we have to prove
$\sum \frac{z^2}{y^2+5xz} \ge \frac {1}{2}$
By Cauchy-Schwarz inequality:
$\sum \frac{z^2}{y^2+5xz}= \sum \frac{z^4}{y^2z^2+5xz^3} \ge \frac{(x^2+y^2+z^2)^2}{\sum x^2y^2 +5 \sum x^3y}$
It suffits to prove
${ 2(x^2+y^2+z^2)^2 \ge \sum x^2y^2 +5 \sum x^3y}$ ,
which is true by your inequality $(x^2+y^2+z^2)^2 \ge 3(x^3y+y^3z+z^3x)$ and the inequality $(x^2+y^2+z^2)^2 \ge 3(x^2y^2+y^2z^2+z^2x^2)$ .

In your last step
Vasc inequality and AM-GM inequality.

$\sum_{cyc}x^4+\sum_{cyc}x^2y^2\geq 2\sum_{cyc}x^3y$

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769 posts

#29 h Apr 7, 2018, 2:11 AM • 2 Y

Y by Adventure10, Mango247

nayel wrote:

Actually if anyone saw my link above it shows that the following one holds for all $a,b,c,m,n$ satisfying $abc = 1$
$\sum_{cyc}\frac {1}{b(ma + nb)}\ge\frac {3}{m + n}$ And it's easily proven using only Cauchy-Schwarz inequality.

Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ , then by Titu

$LHS=\sum_{cyc}\frac{z^2}{ny^2+mxz}\geq \frac{(\sum_{cyc}x^2)^2}{n\sum_{cyc}x^2y^2+m\sum_{cyc}x^3y}$

So we have to show that

$(m+n)(\sum_{cyc}x^2)^2\geq 3n\sum_{cyc}x^2y^2+3m\sum_{cyc}x^3y$

By Rearrangement inequality,

$n(\sum_{cyc}x^2)^2\geq 3n\sum_{cyc}x^2y^2$

By Vasc inequality,

$m(\sum_{cyc}x^2)^2\geq 3m\sum_{cyc}x^3y$

So we're done.

This post has been edited 1 time. Last edited by Takeya.O, Apr 7, 2018, 2:15 AM
Reason: adding explanation

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555 posts

#31 h May 4, 2020, 1:48 PM • 3 Y

Y by Mango247, Mango247, Mango247

nayel wrote:

I think you mean positive reals. :wink:

:wink:

This is very easy, I can't believe it was proposed in a recent olympiad..

Letting $a = \frac xy,b = \frac yz, c = \frac zx$ implies
$LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x}$ Now it remains to prove that
$2(x^2 + y^2 + z^2)^2\ge 3\sum_{cyc}x^2y^2 + 3\sum_{cyc}x^3y$ Which follows by adding the two inequalities
$x^4 + y^4 + z^4\ge x^3y + y^3z + z^3x \\ \\ \sum_{cyc}(x^4 + x^2y^2)\ge \sum_{cyc}2x^3y$ I bet all the contestants solved this one.

EDIT: Here is a related topic: http://www.mathlinks.ro/viewtopic.php?search_id=1111061624&t=165948

How did you come to know that we need to substitute a=x/y at the first place ??

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1013 posts

#32 h Nov 18, 2021, 1:30 PM

Y by

Another one is very nice: a,b,c>0: abc=1 then $\frac{a}{b(a+c)}+\frac{b}{c(b+a)}+\frac{c}{a(c+b)}\ge\frac{3(ab+bc+ca)}{2(a+b+c)}$

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#33 h Dec 31, 2021, 5:43 AM

Y by

sol incorrect

This post has been edited 1 time. Last edited by Albert123, Jul 29, 2022, 12:26 AM

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439 posts

#34 h Dec 31, 2021, 7:26 AM

Y by

Albert123 wrote:

The inequality is equivalent:
$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq \frac{3}{2}$
we have by AM-AG in LHS:
$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq 3\sqrt[3]{\frac{1}{abc(a+b)(b+c)(c+a)}}$
Now, $(a+b)(b+c)(c+a) \geq 8$
$\implies \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq 3\sqrt[3]{\frac{1}{(8)}}$
$\implies \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq \frac{3}{2}$
$\implies \sum_{cyc}{\frac{1}{b(a+b)}} \ge \frac{3}{2}$ . $\blacksquare$

Wrong solution! Giving to denominator less less value makes fraction higher. So this part
is wrong.

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225 posts

#35 h Jun 21, 2022, 11:01 AM

Y by

Kgxtixigct wrote:

nayel wrote:

I think you mean positive reals. :wink:

:wink:

This is very easy, I can't believe it was proposed in a recent olympiad..

Letting $a = \frac xy,b = \frac yz, c = \frac zx$ implies
$LHS = \sum_{cyc}\frac {x^2}{z^2 + xy}\ge\frac {(x^2 + y^2 + z^2)^2}{x^2y^2 + y^2z^2 + z^2x^2 + x^3y + y^3z + z^3x}$ Now it remains to prove that
$2(x^2 + y^2 + z^2)^2\ge 3\sum_{cyc}x^2y^2 + 3\sum_{cyc}x^3y$ Which follows by adding the two inequalities
$x^4 + y^4 + z^4\ge x^3y + y^3z + z^3x \\ \\ \sum_{cyc}(x^4 + x^2y^2)\ge \sum_{cyc}2x^3y$ I bet all the contestants solved this one.

EDIT: Here is a related topic: http://www.mathlinks.ro/viewtopic.php?search_id=1111061624&t=165948

How did you come to know that we need to substitute a=x/y at the first place ??

It's a standard technique to do if you have $abc=1$

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#37 h Jan 20, 2023, 8:48 PM • 1 Y

Y by teomihai

Since $abc=1$ we can use the Ravi substitution $a=\frac{x}{y}$ , $b=\frac{y}{z}$ , $c=\frac{z}{x}$ where $x,y,z>0$ . Now the LHS rewrites as $\sum\limits_{cyc} \frac{y^2 z}{x(xz+y^2)}$ , furthermore:
$\sum\limits_{cyc} \frac{y^4z^4}{xy^2z^3(xz+y^2)}\geq \frac{(x^2y^2+y^2z^2+z^2x^2)^2}{\sum\limits_{cyc} x^2y^2z^4 +\sum\limits_{cyc} xy^4z^3}=\frac{\sum\limits_{cyc}x^4y^4+2\sum\limits_{cyc} x^2y^2z^4}{\sum\limits_{cyc}x^2y^2z^4 +\sum\limits_{cyc} xy^4z^3}=F(x,y,z)$ $F(x,y,z)\geq \frac{3}{2} \iff 2\sum\limits_{cyc}x^4y^4+4\sum\limits_{cyc} x^2y^2z^4\geq 3\sum\limits_{cyc}x^2y^2z^4 +3\sum\limits_{cyc} xy^4z^3$ $\iff 2\sum\limits_{cyc} x^4y^4+\sum\limits_{cyc} x^2y^2z^4\geq 3\sum\limits_{cyc} xy^4z^3\quad (\star)$ Note that AM-GM implies that $\sum\limits_{cyc} y^4z^4+\sum\limits_{cyc} x^2y^4z^2\geq 2\sum\limits_{cyc} xy^4z^3\quad (1)$ and also that $4\sum\limits_{cyc} x^4y^4=\sum\limits_{cyc} (x^4y^4+y^4z^4+y^4z^4+y^4z^4)\geq \sum\limits_{cyc} 4xy^4z^3=4\sum\limits_{cyc} xy^4z^3\Longrightarrow \sum\limits_{cyc} x^4y^4\geq \sum\limits_{cyc} xy^4z^3\quad (2)$ Summing $(1)$ and $(2)$ we get the inequality in $(\star)$ , so we're done.

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#38 h Jan 20, 2023, 9:16 PM

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Marinchoo wrote:

Since $abc=1$ we can use the Ravi substitution $a=\frac{x}{y}$ , $b=\frac{y}{z}$ , $c=\frac{z}{x}$ where $x,y,z>0$ . Now the LHS rewrites as $\sum\limits_{cyc} \frac{y^2 z}{x(xz+y^2)}$ , furthermore:
$\sum\limits_{cyc} \frac{y^4z^4}{xy^2z^3(xz+y^2)}\geq \frac{(x^2y^2+y^2z^2+z^2x^2)^2}{\sum\limits_{cyc} x^2y^2z^4 +\sum\limits_{cyc} xy^4z^3}=\frac{\sum\limits_{cyc}x^4y^4+2\sum\limits_{cyc} x^2y^2z^4}{\sum\limits_{cyc}x^2y^2z^4 +\sum\limits_{cyc} xy^4z^3}=F(x,y,z)$ $F(x,y,z)\geq \frac{3}{2} \iff 2\sum\limits_{cyc}x^4y^4+4\sum\limits_{cyc} x^2y^2z^4\geq 3\sum\limits_{cyc}x^2y^2z^4 +3\sum\limits_{cyc} xy^4z^3$ $\iff 2\sum\limits_{cyc} x^4y^4+\sum\limits_{cyc} x^2y^2z^4\geq 3\sum\limits_{cyc} xy^4z^3\quad (\star)$ Note that AM-GM implies that $\sum\limits_{cyc} y^4z^4+\sum\limits_{cyc} x^2y^4z^2\geq 2\sum\limits_{cyc} xy^4z^3\quad (1)$ and also that $4\sum\limits_{cyc} x^4y^4=\sum\limits_{cyc} (x^4y^4+y^4z^4+y^4z^4+y^4z^4)\geq \sum\limits_{cyc} 4xy^4z^3=4\sum\limits_{cyc} xy^4z^3\Longrightarrow \sum\limits_{cyc} x^4y^4\geq \sum\limits_{cyc} xy^4z^3\quad (2)$ Summing $(1)$ and $(2)$ we get the inequality in $(\star)$ , so we're done.

wow! solution!

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