IMO-1983-6

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sm.joty
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IMO-1983-6

Unread post by sm.joty » Mon Mar 12, 2012 1:17 pm

Let $a,b,c$ be the length of the sides of a triangle.Prove that,
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0$
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sm.joty
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Re: IMO-1983-6

Unread post by sm.joty » Mon Mar 12, 2012 1:55 pm

Hey,I know the official solve but I have an another one.
What about this solution. Is there any mistakes :?: :ugeek:
WLOG we assume that, $a\geq b\geq c$
now clearly we just need to prove,
$a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$
$\Rightarrow \frac{a(a-b)}{c}+\frac{b(b-c)}{a} \geq \frac{c(c-a)}{b}$

now, $\frac{1}{a}\leq\frac{1}{b}\leq\frac{1}{c}$
and
$-a(a-b)\leq 0$
$-b(b-c)\leq 0$
$c(c-a)\leq 0$
by rearrangement inequality,
$\frac{-a(a-b)}{c}+\frac{-b(b-c)}{a}+\frac{c(c-a)}{b}\leq \frac{-a(a-b)}{a}+\frac{-b(b-c)}{b}+\frac{c(c-a)}{c}$
$\frac{-a(a-b)}{c}+\frac{-b(b-c)}{a}+\frac{c(c-a)}{b}\leq 2(c-a)$
$\frac{a(a-b)}{c}+\frac{b(b-c)}{a}-\frac{c(c-a)}{b}\geq -2(c-a) \geq 0$
$\Rightarrow \frac{a(a-b)}{c}+\frac{b(b-c)}{a} \geq \frac{c(c-a)}{b}$
because
$\Rightarrow -2(c-a) \geq 0$
$\Rightarrow a \geq c$
which is obviously true.
So we're done.
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

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nafistiham
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Re: IMO-1983-6

Unread post by nafistiham » Mon Mar 12, 2012 2:24 pm

sm.joty wrote:WLOG we assume that, $a\geq b\geq c$
now clearly we just need to prove,
$a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$
why that ? is
\[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0\]
and
\[a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)\]
equal ?
shouldn't it be
\[a^2b(a-b)+b^2c(b-c) \geq -c^2a(c-a)\]
probably I am a little confused :?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: IMO-1983-6

Unread post by nafistiham » Mon Mar 12, 2012 2:32 pm

and again, isn't it just a one liner muirhead ?
as we can see
\[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq0\]
\[\Rightarrow a^3b-a^2b^2+b^3c-b^2c^2+c^3a-c^2a^2\geq0\]
\[\Rightarrow a^3b+b^3c+c^3a\geq a^2b^2+b^2c^2+c^2a^2\]
\[\Rightarrow [3,1]\geq [2,2]\]
Last edited by nafistiham on Mon Mar 12, 2012 2:43 pm, edited 2 times in total.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
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CSE Dept. SUST -HSC 14'
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nafistiham@gmail

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sm.joty
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Re: IMO-1983-6

Unread post by sm.joty » Mon Mar 12, 2012 2:39 pm

nafistiham wrote:
sm.joty wrote:WLOG we assume that, $a\geq b\geq c$
now clearly we just need to prove,
$a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$
why that ? is
\[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0\]
and
\[a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)\]
equal ?
shouldn't it be
\[a^2b(a-b)+b^2c(b-c) \geq -c^2a(c-a)\]
probably I am a little confused :?
Tiham,
for
$x,y,z$
if $x+y-c \geq 0$ then must $x+y+c \geq 0$

on the other hand, according to my assumption here $a^2b(a-b)$ and $b^2c(b-c)$ is positive.
But $c^2a(c-a)$ is negative,
thus if $a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$ then obiously
$a^2b(a-b)+b^2c(b-c)+ c^2a(c-a) \geq 0$
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

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sm.joty
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Re: IMO-1983-6

Unread post by sm.joty » Mon Mar 12, 2012 2:42 pm

nafistiham wrote:
sm.joty wrote:WLOG we assume that, $a\geq b\geq c$
now clearly we just need to prove,
$a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$
why that ? is
\[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0\]
and
\[a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)\]
equal ?
shouldn't it be
\[a^2b(a-b)+b^2c(b-c) \geq -c^2a(c-a)\]
probably I am a little confused :?
Tiham,
for
$x,y,z$
if $x+y-c \geq 0$ then obiously $x+y+c \geq 0$

on the other hand, according to my assumption here $a^2b(a-b)$ and $b^2c(b-c)$ is positive.
But $c^2a(c-a)$ is negative,
thus if $a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$ then obiously
$a^2b(a-b)+b^2c(b-c)+ c^2a(c-a) \geq 0$
:)
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

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nafistiham
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Re: IMO-1983-6

Unread post by nafistiham » Mon Mar 12, 2012 3:09 pm

sm.joty wrote:
nafistiham wrote:
sm.joty wrote:WLOG we assume that, $a\geq b\geq c$
now clearly we just need to prove,
$a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$
why that ? is
\[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0\]
and
\[a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)\]
equal ?
shouldn't it be
\[a^2b(a-b)+b^2c(b-c) \geq -c^2a(c-a)\]
probably I am a little confused :?
Tiham,
for
$x,y,z$
if $x+y-c \geq 0$ then obiously $x+y+c \geq 0$

on the other hand, according to my assumption here $a^2b(a-b)$ and $b^2c(b-c)$ is positive.
But $c^2a(c-a)$ is negative,
thus if $a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$ then obiously
$a^2b(a-b)+b^2c(b-c)+ c^2a(c-a) \geq 0$
:)
thanks for clarifying. :oops: :oops:
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

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*Mahi*
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Re: IMO-1983-6

Unread post by *Mahi* » Mon Mar 12, 2012 3:31 pm

sm.joty wrote: now, $\frac{1}{a}\leq\frac{1}{b}\leq\frac{1}{c}$
and
$-a(a-b)\leq 0$
$-b(b-c)\leq 0$
$c(c-a)\leq 0$
by rearrangement inequality,
$\frac{-a(a-b)}{c}+\frac{-b(b-c)}{a}+\frac{c(c-a)}{b}\leq \frac{-a(a-b)}{a}+\frac{-b(b-c)}{b}+\frac{c(c-a)}{c}$
This is not true. Rearrangement inequality implies $a\geq b \ge c$ and $x \ge y \ge z \Rightarrow$
$ax+by+cz \ge ay+bz+cx$ and so on. You have to prove $c(c-a) \ge -b(b-c) \ge -a(a-b)$ for what you said.
Please read Forum Guide and Rules before you post.

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Corei13
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Re: IMO-1983-6

Unread post by Corei13 » Mon Mar 12, 2012 8:03 pm

পোলাপাইন, This is not a symmetric inequality but a cyclic one. So you Can't WLOG assume that $a\ge b\ge c$, instead you can assume WLOG either $a\ge b \ge c$ OR $a\le b\le x$. And, as it is not symmetric, you can't use Muirhead 's inequality.
ধনঞ্জয় বিশ্বাস

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Re: IMO-1983-6

Unread post by Sazid Akhter Turzo » Mon Mar 12, 2012 9:20 pm

এইটাই কেউ পার না । Incircle-টা যে যে অংশে a,b ও c-কে ভাগ করে, তা চিন্তা কর । এরপর Cauchy-Schwarz use কর, the problem is now solved!

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