Any other solution to this inequality ?

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January 14th 2011, 05:22 AM
Tarask

Any other solution to this inequality ?

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:
http://latex.codecogs.com/gif.latex?...c{9}{4(a+b+c)}
And please , don't tell me it is pre-calculus ! (Giggle)

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

Moderator edit: Moved from Pre-algebra and Algebra. Closed until OP reads and follows the subforum rules for posting here.
January 14th 2011, 07:08 AM
Pranas

Maybe it's a matter of notation in different regions, bet personally I am not exactly sure of summation
$\displaystyle \[\sum\limits_{cyc} {} \]$
January 14th 2011, 07:18 AM
Also sprach Zarathustra

Quote:

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
January 14th 2011, 07:20 AM
Tarask

Can i give a hint ?
January 14th 2011, 08:38 AM
emakarov

I still don't understand what $\displaystyle\sum_{cyc}\frac{a}{(b+c)^2}$ is. Is it $\displaystyle\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\ frac{c}{(a+b)^2}$ ?
January 14th 2011, 11:36 AM
Tarask

Yes it is !
January 14th 2011, 12:58 PM
Tarask

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

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