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Home » Mathematics » High School Pre-Olympiad (Ages 14+)

inequality with ab+bc+ca=1
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chien than
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inequality with ab+bc+ca=1

Let $a;b;c>0$ such that $ab+bc+ca=1$
Prove that
$3+3(a^2+b^2+c^2) \geq 9(a^2b^2+b^2c^2+c^2a^2)+81a^2b^2c^2$

Posted: Fri Oct 09, 2009 2:25 am

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TRAN THAI HUNG
Riemann Hypothesis

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Re: inequality with ab+bc+ca=1

chien than wrote:

Let $a;b;c > 0$ such that $ab + bc + ca = 1$
Prove that
$3 + 3(a^2 + b^2 + c^2) \geq 9(a^2b^2 + b^2c^2 + c^2a^2) + 81a^2b^2c^2$

I think we can normalize it and then use S.O.S

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Posted: Fri Oct 09, 2009 9:03 am

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jingjun
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Re: inequality with ab+bc+ca=1

chien than wrote:

Let $a;b;c > 0$ such that $ab + bc + ca = 1$
Prove that
$3 + 3(a^2 + b^2 + c^2) \geq 9(a^2b^2 + b^2c^2 + c^2a^2) + 81a^2b^2c^2$

let $a + b + c = p,ab + bc + ca = \frac {p^{2} - q^{2}}{3},abc = r ,$ $and \frac {p^{2} - q^{2}}{3} = 1$

so it's equivalent to proof $f(p,q,r) = - 12 - 81r^{2} + 3p^{2} + 18pr\geq 0.$

if $p\geq 2, f(p,q,r)\geq f(p,q,0) = 3(p - 2)^{2}\geq 0.$

if $2\geq p\geq \sqrt{3}, f(p,q,r)\geq f(p,p^{2} - 3,\frac {p^{3} - 3p(p^{2} - 3) - 2(p^{2} - 3)^{\frac {3}{2}})}{27})$

$= \frac {4p(p^{2} - 3)(4p + 2p^{2} - 1)(2 - p) + 2(1 - (p^{2} - 3)^{\frac {3}{2}})}{9}\geq 0.$

Posted: Sat Nov 07, 2009 9:30 pm

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