Simple Inequality Problem

matheas

let $a,b,c\ge 0$ such that $ab + bc + ca = 1$

prove that

$\frac {a}{bc + 1} + \frac {b}{ac + 1} + \frac {c}{ab + 1}\ge \frac {3\sqrt{3}}{4}$

**Posted:** Sat Oct 03, 2009 1:21 am

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FelixD

We have $\sum \frac{a}{bc+1} = \sum a - abc\sum \frac{1}{bc+1} \ge \sqrt{3}-abc\sum \frac{\sqrt[4]{27}}{4\sqrt[4]{bc}}$ . Thus it is left to prove
$\frac{1}{\sqrt[4]{3}} \ge \sum x^4y^3z^3,$
where $x^4y^4+y^4z^4+z^4x^4=1$ . (hence $a=x^4$ and so on)
This can be easily done by AmGm.

**Posted:** Sat Oct 03, 2009 2:00 am

Tourish · Hodge Conjecture Offline Joined: 11 Sep 2009 Posts: 62

**Posted:** Sat Oct 03, 2009 7:19 am

akashram

USe titu's lemma and then apply amgm to the smplified form

**Posted:** Fri Nov 06, 2009 3:07 am

geniusbliss

@akashram
please post your solution instead of claiming.

**Posted:** Fri Nov 06, 2009 6:57 pm

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Agr_94_Math · Yang-Mills Theory Online Joined: 17 Feb 2008 Posts: 722

For $a,b,c$ positive reals, $a+b+c=1$ , the following inequality holds:
$\sum \frac{a}{bc+1} \ge \frac{9}{10}$ .

**Posted:** Fri Nov 06, 2009 9:29 pm

Dimitris X

**Posted:** Sat Nov 07, 2009 1:06 am

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**Posted:** Sat Nov 07, 2009 5:18 pm

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Agr_94_Math · Yang-Mills Theory Online Joined: 17 Feb 2008 Posts: 722

Another proof using the generalized Means Inequality:
Since $a+b+c=1$ , assigning the weights $a,b,c$ to the expressions $\frac{1}{bc+1}, \frac{1}{ca+1}, \frac{1}{ab+1}$ respectively, using weighted AM-HM, it is done.
That is $M_1(a) \ge M_{-1} (a)$ .

**Posted:** Sat Nov 07, 2009 6:17 pm

spanferkel · Yang-Mills Theory Offline Joined: 15 Oct 2005 Posts: 661

**Posted:** Sun Nov 08, 2009 7:57 am

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