## IMO-1983-6

Discussion on International Mathematical Olympiad (IMO)
sm.joty
Posts: 327
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Location: Dhaka

### IMO-1983-6

Let $a,b,c$ be the length of the sides of a triangle.Prove that,
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0$
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

sm.joty
Posts: 327
Joined: Thu Aug 18, 2011 12:42 am
Location: Dhaka

### Re: IMO-1983-6

Hey,I know the official solve but I have an another one.
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

nafistiham
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### Re: IMO-1983-6

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
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

nafistiham
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### Re: IMO-1983-6

and again, isn't it just a one liner muirhead ?
as we can see
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.

sm.joty
Posts: 327
Joined: Thu Aug 18, 2011 12:42 am
Location: Dhaka

### Re: IMO-1983-6

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.........

sm.joty
Posts: 327
Joined: Thu Aug 18, 2011 12:42 am
Location: Dhaka

### Re: IMO-1983-6

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.........

nafistiham
Posts: 829
Joined: Mon Oct 17, 2011 3:56 pm
Location: 24.758613,90.400161
Contact:

### Re: IMO-1983-6

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.
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.

*Mahi*
Posts: 1175
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### Re: IMO-1983-6

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.

Use $L^AT_EX$, It makes our work a lot easier!

Corei13
Posts: 153
Joined: Tue Dec 07, 2010 9:10 pm
Location: Chittagong

### Re: IMO-1983-6

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

Sazid Akhter Turzo
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### Re: IMO-1983-6

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!