Nobody could solve this?

nice problem.My solution We have: $ \frac {a_ia_j}{c_i + c_j} = \int_{0}^{1} a_ia_jx^{c_i + c_j - 1}dt$ So:$ \sum_{i,j = 1}^{n}\frac {a_{i}a_{j}}{c_{i} + c_{j}} = \int_{0}^{1}\sum_{i,j = 1}^{n}a_ia_jt^{c_i + c_j - 1}dt = \int_{0}^{1}(\sum_{i = 1}^{n}a_it^{c_i - \frac {1}{2}})^2dt$ Similary:$ \sum_{i,j = 1}^{n}\frac {b_{i}b_{j}}{c_{i} + c_{j}} = \int_{0}^{1}(\sum_{i = 1}^{n}b_it^{c_i - \frac1{2}})^2$ Let $ f(x) = \sum_{i = 1}^{n}a_ix^{c_i - \frac {1}{2}}; g(x) = \sum_{i = 1}^{n}b_it^{c_i - \frac1{2}}$ We have:$ \sum_{i,j = 1}^{n}\frac {a_{i}b_{j}}{c_{i} + c_{j}} = \int_{0}^{1}a_ib_jt^{c_i + c_j - 1} = \int_{0}^{1}f(t)g(t)dt$ Using Ineq Cauchy_Schwarz we have: $ \int_{0}^{1}(f(x))^2dx.\int_{0}^{1}(g(x))^2dx\geq (\int_{0}^{1}f(x)g(x)dx)^2$ This is necessary to prove

Hm, right! Unfortunately in Bulgaria we don't learn integrals at school so this occurs to be a hard inequality. This is one of the solutions. There are other two (known by me). The first uses derivatives! The second is "elemntary". By "elementary" I mean that it doesn't uses wide range of theorems and formulas but only a simple things. However it is very hard to manage to find this solution. And the solutions with Cauchy-Schwartz is the easiest but as I told you we don't learn it in this form at school. Actually the highest result on this problem was made of a girl and it was 6 points of 7. The next was 4 of 7 and it was done by a russian guest.

I tried, but I can not find a solution other I really hope to exist a nice solution (Elementary) because I think looking for a good idea is a difficult task Can you post full problem Bulgaria MO,TSR 2008,2009?

Hi, Can you post the other solution please? Thanks a lot.

math10 wrote: $ \int_{0}^{1}\sum_{i,j = 1}^{n}a_ia_jt^{c_i + c_j - 1}dt = \int_{0}^{1}(\sum_{i = 1}^{n}a_it^{c_i - \frac {1}{2}})^2dt$ Sorry but why is this true?