IMO-1983-6
Posted: Mon Mar 12, 2012 1:17 pm
Let $a,b,c$ be the length of the sides of a triangle.Prove that,
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0$
$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0$
why that ? issm.joty wrote:WLOG we assume that, $a\geq b\geq c$
now clearly we just need to prove,
$a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$
Tiham,nafistiham wrote:why that ? issm.joty wrote:WLOG we assume that, $a\geq b\geq c$
now clearly we just need to prove,
$a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$
\[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0\]
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,nafistiham wrote:why that ? issm.joty wrote:WLOG we assume that, $a\geq b\geq c$
now clearly we just need to prove,
$a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$
\[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0\]
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
thanks for clarifying.sm.joty wrote:Tiham,nafistiham wrote:why that ? issm.joty wrote:WLOG we assume that, $a\geq b\geq c$
now clearly we just need to prove,
$a^2b(a-b)+b^2c(b-c) \geq c^2a(c-a)$
\[a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\geq 0\]
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
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
$a^2b(a-b)+b^2c(b-c)+ c^2a(c-a) \geq 0$
This is not true. Rearrangement inequality implies $a\geq b \ge c$ and $x \ge y \ge z \Rightarrow$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,
$\frac{-a(a-b)}{c}+\frac{-b(b-c)}{a}+\frac{c(c-a)}{b}\leq \frac{-a(a-b)}{a}+\frac{-b(b-c)}{b}+\frac{c(c-a)}{c}$