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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
1 viewing
jlacosta
Mar 2, 2025
0 replies
I think I know why this problem was rejected by IMO PSC several times...
mshtand1   0
4 minutes ago
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 11.8
Exactly $102$ country leaders arrived at the IMO. At the final session, the IMO chairperson wants to introduce some changes to the regulations, which the leaders must approve. To pass the changes, the chairperson must gather at least \(\frac{2}{3}\) of the votes "FOR" out of the total number of leaders. Some leaders do not attend such meetings, and it is known that there will be exactly $81$ leaders present. The chairperson must seat them in a square-shaped conference hall of size \(9 \times 9\), where each leader will be seated in a designated \(1 \times 1\) cell. It is known that exactly $28$ of these $81$ leaders will surely support the chairperson, i.e., they will always vote "FOR." All others will vote as follows: At the last second of voting, they will look at how their neighbors voted up to that moment — neighbors are defined as leaders seated in adjacent cells \(1 \times 1\) (sharing a side). If the majority of neighbors voted "FOR," they will also vote "FOR." If there is no such majority, they will vote "AGAINST." For example, a leader seated in a corner of the hall has exactly $2$ neighbors and will vote "FOR" only if both of their neighbors voted "FOR."

(a) Can the IMO chairperson arrange their $28$ supporters so that they vote "FOR" in the first second of voting and thereby secure a "FOR" vote from at least \(\frac{2}{3}\) of all $102$ leaders?

(b) What is the maximum number of "FOR" votes the chairperson can obtain by seating their 28 supporters appropriately?

Proposed by Bogdan Rublov
0 replies
mshtand1
4 minutes ago
0 replies
It was inevitable: Maestro's problem on UKR MO
mshtand1   1
N 12 minutes ago by RagvaloD
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 10.6
Find all triples of nonnegative real numbers \((x, y, z)\) that satisfy the equality:
\[
\frac{\left(x^2 - y\right)(1 - y)}{(x - y)^2} + \frac{\left(y^2 - z\right)(1 - z)}{(y - z)^2} + \frac{\left(z^2 - x\right)(1 - x)}{(z - x)^2} = 3
\]Proposed by Vadym Solomka
1 reply
mshtand1
an hour ago
RagvaloD
12 minutes ago
This problem has unintended solution, found by almost all who solved it :(
mshtand1   0
17 minutes ago
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 11.7
Given a triangle \(ABC\), an arbitrary point \(D\) is chosen on the side \(AC\). In triangles \(ABD\) and \(CBD\), the angle bisectors \(BK\) and \(BL\) are drawn, respectively. The point \(O\) is the circumcenter of \(\triangle KBL\). Prove that the second intersection point of the circumcircles of triangles \(ABL\) and \(CBK\) lies on the line \(OD\).

Proposed by Anton Trygub
0 replies
mshtand1
17 minutes ago
0 replies
Best geo on this year UKR MO, even contestants call it marvellous
mshtand1   1
N 20 minutes ago by khina
Source: Ukrainian Mathematical Olympiad 2025. Day 2, Problem 10.8
In an acute-angled triangle \(ABC\) \((AC > BC)\) with altitude \(AD\), the following points are marked: \(H\) - the orthocenter, \(O\) - the circumcenter, \(K\) - the midpoint of side \(AB\). Inside the triangle \(\triangle ADC\), there is a point \(P\) such that the following equality holds:
\[
\angle KPD + \angle ACB = 2 \angle OPH = 180^{\circ}
\]Prove that
\[
BH = 2PD
\]
Proposed by Vadym Solomka
1 reply
mshtand1
37 minutes ago
khina
20 minutes ago
No more topics!
Three variables, cyclic
PhilAndrew   33
N Jan 20, 2023 by teomihai
Source: Zhautykov Olympiad, Kazakhstan 2008, Question 6
Let $ a, b, c$ be positive integers for which $ abc = 1$. Prove that
$ \sum \frac{1}{b(a+b)} \ge \frac{3}{2}$.
33 replies
PhilAndrew
Jan 18, 2008
teomihai
Jan 20, 2023
Three variables, cyclic
G H J
Source: Zhautykov Olympiad, Kazakhstan 2008, Question 6
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PhilAndrew
207 posts
#1 • 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}$.
Z K Y
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nayel
1394 posts
#2 • 3 Y
Y by vsathiam, Adventure10, Mango247
I think you mean positive reals. :wink:

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

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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pardesi
3183 posts
#3 • 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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Erken
1363 posts
#4 • 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 :wink:
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pardesi
3183 posts
#5 • 2 Y
Y by Adventure10, Mango247
sorry i had copied the inequality wrong :rotfl:
sorry nayel :(
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pardesi
3183 posts
#6 • 2 Y
Y by Adventure10, Mango247
and how is that from cauchy
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Erken
1363 posts
#7 • 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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pardesi
3183 posts
#8 • 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. :o
u have to multiply by $ x^{2}$
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Erken
1363 posts
#9 • 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. :o
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 :roll:
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pardesi
3183 posts
#10 • 2 Y
Y by Adventure10, Mango247
congrats anyways :D
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nayel
1394 posts
#11 • 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. :o
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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pardesi
3183 posts
#12 • 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 :wink:
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nayel
1394 posts
#13 • 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 :wink:

That's what makes the problem a bit more standard. Otherwise it would be trivial right? :roll:
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pardesi
3183 posts
#14 • 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 :o
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nayel
1394 posts
#15 • 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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lasha
204 posts
#16 • 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
279 posts
#17 • 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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Vasc
2861 posts
#18 • 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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ehku
253 posts
#19 • 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
1394 posts
#20 • 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. :P
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Uzbekistan
560 posts
#21 • 2 Y
Y by Adventure10, Mango247
EAsy TRY REARRANGEMENT ineq
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henderson
312 posts
#22 • 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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szl6208
2032 posts
#23 • 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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arqady
30146 posts
#24 • 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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cucugauss
8 posts
#26 • 1 Y
Y by Adventure10
nayel wrote:
I think you mean positive reals. :wink:

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

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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Takeya.O
769 posts
#28 • 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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Takeya.O
769 posts
#29 • 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. :P

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. :D
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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Kgxtixigct
555 posts
#31 • 3 Y
Y by Mango247, Mango247, Mango247
nayel wrote:
I think you mean positive reals. :wink:

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

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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jokehim
1013 posts
#32
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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Albert123
204 posts
#33
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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Jalil_Huseynov
439 posts
#34
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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minusonetwelth
225 posts
#35
Y by
Kgxtixigct wrote:
nayel wrote:
I think you mean positive reals. :wink:

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

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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Marinchoo
407 posts
#37 • 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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teomihai
2971 posts
#38
Y by
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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