Any other solution to this inequality ?

**Tarask** · January 14th 2011, 05:22 AM

Hi everyone !
I found that inequality that i proved , but i wanted another solution .
Show that for all a,b and c positive real numbers , the following inequality holds:
$\sum_{cyc}\frac{a}{(b+c)^2}\geq \frac{9}{4(a+b+c)}$
And please , don't tell me it is pre-calculus !

Anybody with another solution ?
(if you need mine , say it

)

Moderator edit: Moved from Pre-algebra and Algebra. Closed until OP reads and follows the subforum rules for posting here.

**Pranas** · January 14th 2011, 07:08 AM

Maybe it's a matter of notation in different regions, bet personally I am not exactly sure of summation
$\displaystyle \[\sum\limits_{cyc} {} \]$

**Also sprach Zarathustra** · January 14th 2011, 07:18 AM

Originally Posted by Pranas

Maybe it's a matter of notation in different regions, bet personally I am not exactly sure of summation
$\displaystyle \[\sum\limits_{cyc} {} \]$

cyc=cyclic

**Tarask** · January 14th 2011, 07:20 AM

Can i give a hint ?

**emakarov** · January 14th 2011, 08:38 AM

I still don't understand what $\displaystyle\sum_{cyc}\frac{a}{(b+c)^2}$ is. Is it $\displaystyle\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\ frac{c}{(a+b)^2}$ ?

**Tarask** · January 14th 2011, 11:36 AM

Yes it is !

**Tarask** · January 14th 2011, 12:58 PM

Well , you must use both Cauchy-Schwartz and Nesbitt inequalities .

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