# Thread: Any other solution to this inequality ?

1. ## 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:
$\sum_{cyc}\frac{a}{(b+c)^2}\geq \frac{9}{4(a+b+c)}$
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.

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

3. 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

4. Can i give a hint ?

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

6. Yes it is !

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