mr.danh wrote: Prove the inequality $ \sum_{i < j} \frac {a_ia_j}{a_i + a_j} \le \frac {n}{2(a_1 + a_2 + ... + a_n)}\sum_{i < j} a_ia_j$ for all positive real numbers $ a_1, a_2, . . . , a_n$. The general one can be proved by the same way as the case $ n=3$, . \[ \frac{{ab}}{{a + b}} + \frac{{bc}}{{b + c}} + \frac{{ca}}{{c + a}} \le \frac{{3\left( {ab + bc + ca} \right)}}{{2\left( {a + b + c} \right)}} \] And to proved this inequality we use this $ \frac{1}{a} + \frac{1}{b} \ge \frac{4}{{a + b}}$, .

Honey_S wrote: And to proved this inequality we use this $ \frac {1}{a} + \frac {1}{b} \ge \frac {4}{{a + b}}$, . I think you're wrong...

BaBaK Ghalebi wrote: Honey_S wrote: And to proved this inequality we use this $ \frac {1}{a} + \frac {1}{b} \ge \frac {4}{{a + b}}$, . I think you're wrong... Of course ... not, my friend, . To make it more clearly for you \[ 2\left( {\sum a } \right)\left( {\sum {\frac{{ab}}{{a + b}}} } \right) = 2\sum {ab} + 2abc\sum {\frac{1}{{a + b}}} \le 3\sum {ab} \] Besides that, thanks you for a look at my solution here http://www.mathlinks.ro/viewtopic.php?t=224946&sid=2905796b9149d15b146dcde414121b86

yeah,you're right... just a mistyped in your last post,it should be $ 3\sum ab$ instead of $ 3abc$.

IMOSL 2006