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orl
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Post Posted: Nov 09, 2005, 2:16 pm • # 1 


Let a, b, c be positive real numbers such that abc = 1. Prove that
\frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}...

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Last edited by orl on Aug 10, 2008, 8:54 am, edited 1 time in total.
 
 
darij grinberg
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Post Posted: Nov 09, 2005, 2:27 pm • # 2 


Stronger inequality at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25778 .

Darij

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Valentin Vornicu
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Post Posted: Nov 09, 2005, 11:42 pm • # 3 


From Cauchy we have \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \Big( a(b+c)+b(c+a)+c(a+b) \Big) \geq \left( \fr... and as \frac 1a + \frac 1b + \frac 1c = ab+bc+ca we obtain that \left( \frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)}\right) \geq \frac{ ab+bc+ca} 2 \geq \frac{\sqrt [3]{a^2... which is what we wanted to prove.

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babylearnmath294
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Post Posted: Nov 12, 2005, 10:46 pm • # 4 


We have abc=1

Therefore 1/(a^3)(b+c) = bc/(a^2)(b+c)

From Cauchy we have :


bc/(a^2)(b+c) + 1/4b + 1/4c = bc/(a^2)(b+c)+(b+c)/4bc >= 1/a

Similarly we have P>= 1/2a + 1/2b +1/2c>=3/2

which is what we wanted to prove
 
 
campos
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Post Posted: Jun 28, 2006, 9:46 pm • # 5 


hope this has not been forgotten! :D i came to a weird solution, but cool and nice!

we want to prove there exists and r such that
\frac{1}{a^3(b+c)}\geq \frac{3}{2}\frac{a^r}{a^r+b^r+c^r}.

take b=c and use that a=b^{-2}, then we get to

\frac{b^6}{2b}\geq \frac{3}{2}\frac{b^{-2r}}{b^{-2r}+2b^r} or b^5\geq \frac{3}{1+2b^{3r}} or f(b)=2b^{3r+5}+b^5-3\geq 0.

Obviously, f(1)=0, so we want f'(1)=0, this is
f'(b)=2(3r+5)b^{3r+2}+5b^4, so we want 2(3r+5)+5=0, or r=\frac{-5}{2}.

we want to prove that
\frac{1}{a^3(b+c)}\geq \frac{3}{2}\frac{a^\frac{-5}{2}}{a^\frac{-5}{2}+b^\frac{-5}{2}+c^\frac{-5}{2}}.

use that a=\frac{1}{bc} and let x=\sqrt{b} and y=\sqrt{c}.

we want to prove that
\frac{x^6y^6}{x^2+y^2}\geq \frac{3}{2}\frac{x^5y^5}{x^5y^5+\frac{1}{x^5}+\frac{1}{y^5}} or (after a bit of simplifications)

2(x^{10}y^{10}+x^5+y^5)\geq 3x^4y^4(x^2+y^2)

from am-gm and rearrangement we have that

2(x^{10}y^{10}+x^5+y^5)\geq (x^{10}y^{10}+2x^4y)+(x^{10}y^{10}+2xy^4)\geq 3x^4y^4(x^2+y^2) and we're done!
 
 
me@home
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Post Posted: Jan 28, 2007, 6:21 pm • # 6 


Sorry for bringing this up again but I just recently did this problem from the inequalities packet
 
 
gollywog
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Post Posted: Feb 11, 2007, 5:50 pm • # 7 


the proof is long and i'm too tired to rewrite it so i'll just note:

use well known substition a = \frac{x}{y} and analogically for b and c, so the assumption that abc = 1 is satisfied, ok

now we multiply everything, we have something like this
\sum \frac{(y^{3}x)(y^{3}z)}{(x^{3}z)(y^{3}z)+(x^{3}y)(y^{3}x)} then we prove it pretty same as Nesbitt

soz lads... i thought this way worths a mention...

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x^n y^n=z^n
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Post Posted: Jul 10, 2007, 12:40 am • # 8 


orl wrote:
Let a,b,c be positive real numbers such that abc=1.


Valentin Vornicu wrote:
\frac{\sqrt [3]{a^{2}b^{2}c^{2}}}{2}= \frac{3}{2}


:huh:?
 
 
kunny
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Post Posted: Jul 10, 2007, 12:56 am • # 9 


x^n+y^n=z^n wrote:
orl wrote:
Let a,b,c be positive real numbers such that abc=1.


Valentin Vornicu wrote:
\frac{\sqrt [3]{a^{2}b^{2}c^{2}}}{2}= \frac{3}{2}


:huh:?


Valentin seemed to have made a typo, it should be \frac{3}{2}\sqrt [3]{a^{2}b^{2}c^{2}}}= \frac{3}{2}.
 
 
SnowEverywhere
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Post Posted: Mar 20, 2010, 2:01 pm • # 10 


Hello. I am really sorry for bringing up such an old topic. I am really new to olympiad inequalities and was just curious as to how you guys would have come up with that solution with Cauchy.

Did you instinctively know that you were looking for ab + bc + ac in some form and know that the expression a(b+c) + b(a+c) + c(a+b) would similarly give you this expression?

Or did you narrow down the possible approaches (from C-S, AM-GM, finding some expression always \ge 3/2 but \le the left hand side, etc.) and try applying them until something worked?
 
 
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