it is very easy problem just use Holder and u are done!!!!

janni wrote: it is very easy problem just use Holder and u are done!!!! i will be happy to see your solution.

andria wrote: janni wrote: it is very easy problem just use Holder and u are done!!!! i will be happy to see your solution.[/quote i am sorry i couldn't write inquality .......

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is it ok ????? u can read!!!

propossed by Mohammad Ahmadi

Any solutions?

One idea that came to my mind is : induction

any other idea

It suffices to show $\sum_{i<j} \left(\frac{4a_ia_j}{(a_i+a_j)^2}\right)\geq\frac{n^3}{2\sum_{i=1}^{n}a_i}-\frac{n}{2}$ or $\sum_{i<j} \left(\frac{4a_ia_j}{(\frac{a_i}{a_j}+\frac{a_j}{a_i})(a_ia_j+a_ia_j)}\right)\geq\frac{n^3}{2\sum_{i=1}^{n}a_i}-\frac{n}{2}$ Its equal to show $\sum_{i<j} \left(\frac{1}{(\frac{a_i}{a_j}+\frac{a_j}{a_i})}\right)\geq\frac{n^3}{4\sum_{i=1}^{n}a_i}-\frac{n}{4}$ with use T2 we have $ \frac{\binom{n}{2}^2}{(\sum_{i=1}^{n}a_i)(\sum_{i=1}^{n}\frac{1}{a_i})-n}\geq\frac{n^3}{4\sum_{i=1}^{n}a_i}-\frac{n}{4}$ or $ \binom{n}{2}^2\geq\left(\frac{n^3}{4\sum_{i=1}^{n}a_i}-\frac{n}{4}\right)({(\sum_{i=1}^{n}a_i)(n)-n})$ and its easy from $ n^2 +(\sum_{i=1}^{n}a_i)^2 \geq\left(2n\right)({(\sum_{i=1}^{n}a_i)})$ Q.E.D

it's enough to show that $ \sum _ {i<j} \frac { 4 }{(\frac {1}{a_i} +\frac {1}{a_j} ) ( a_i +a_j )} \geq { \frac {n^3}{2 \sum _{i=1}^{n}a_i } - \frac {n}{2} } $ By cauchy we have $ LHS . ( \sum _{i<j} (a_i + a_j)(\frac {1}{a_i} +\frac {1}{a_j}) ) \geq n^2(n-1)^2 $ It's easy to check that $ \sum _{i<j} (a_i + a_j)(\frac {1}{a_i} + \frac {1}{a_j}) =n(n-2) + n \sum _{i<j} a_i $ So $ LHS \geq \frac {n(n-1)^2}{n-2+ \sum _{i=1}^{n} a_i} $ so it's enough to show that $ \frac {n(n-1)^2}{\sum _{i=1}^{n} a_i +n-2 } \geq \frac {n(n^2 - \sum_{i=1}^{n}a_i )}{2\sum_{i=1}^{n}a_i} $ $ \leftrightarrow (\sum _{i=1}^{n} a_i )^2 + (\sum_{i=1}^{n} a_i )(n^2-3n) - (n^3 _ 2n^2) \geq 0 $ and it's obvious that we should just say that $ \sum _{i=1}^{n} a_i \geq n $ and because $ \sum_{i=1}^{n} \frac{1}{a_i} = n $ , we have that and the problem is finished $ \blacksquare $