The prime inequality learning problem

orl

#1

The prime inequality learning problem

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darij grinberg

#2

Re: the prime inequality learning problem

**Posted:** Wed Nov 09, 2005 11:27 pm

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#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.

**Posted:** Thu Nov 10, 2005 8:42 am

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babylearnmath294 · New Member Offline Joined: 12 Nov 2005 Posts: 12

#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

**Posted:** Sun Nov 13, 2005 7:46 am

campos

#5

hope this has not been forgotten! Mr. Green

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!

**Posted:** Thu Jun 29, 2006 6:46 am

me@home

#6

Sorry for bringing this up again but I just recently did this problem from the inequalities packet

**Posted:** Mon Jan 29, 2007 3:21 am

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gollywog

#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...

**Posted:** Mon Feb 12, 2007 2:50 am

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x^n y^n=z^n · New Member Offline Joined: 25 Jul 2006 Posts: 2

#8

**Posted:** Tue Jul 10, 2007 9:40 am

kunny

#9

**Posted:** Tue Jul 10, 2007 9:56 am