Hmm... I've got straightforward solution, I hope there is some niecer We decompose: $(\sum a_i)(\sum b_i)\geq (\sum a_i+b_i)(\sum\frac{a_ib_i}{a_i+b_i}) $ into sum of many "smaller" inequalities which will turn to be true: For each $i$ this ineq is true: $(a_ib_i)\geq (a_i+b_i)\frac{a_ib_i}{a_i+b_i}$ cause it's equivalent to: $a_ib_i\geq a_ib_i$ which is true. For each $i<j$ we have: $a_ib_j+a_jb_i\geq (a_i+b_i)\frac{a_jb_j}{a_j+b_j}+(a_j+b_j)\frac{a_ib_i}{a_i+b_i}$ It's true because it's equivalent to: $\frac{(a_ib_j-a_jb_i)^2}{(a_i+b_i)(a_j+b_j)}\geq 0$ And it's certainly true. Now we have: $(\sum a_i)(\sum b_i)=(\sum a_ib_i)+(\sum_{i<j} a_ib_j+a_jb_i)\geq \\ \geq (\sum (a_i+b_i)\frac{a_ib_i}{a_i+b_i})+(\sum_{i<j} (a_i+b_i)\frac{a_jb_j}{a_j+b_j}+(a_j+b_j)\frac{a_ib_i}{a_i+b_i})=\\=(\sum a_i+b_i)(\sum\frac{a_ib_i}{a_i+b_i}) $

Pascual2005 wrote: 1. Prove the following inequality for positive reals $a_1,a_2...,a_n$ and $b_1,b_2...,b_n$: $(\sum a_i)(\sum b_i)\geq (\sum a_i+b_i)(\sum\frac{a_ib_i}{a_i+b_i})$ In http://www.mathlinks.ro/Forum/viewtopic.php?t=16582 post #9, I proved the inequality $\sum_{i=1}^{n}A_{i} \cdot \sum_{i=1}^{n}B_{i}\geq \sum_{i=1}^{n}\frac{A_{i}B_{i}}{A_{i}+B_{i}}\cdot \left( \sum_{i=1}^{n}A_{i}+\sum_{i=1}^{n}B_{i}\right) $ for positive reals $A_1$, $A_2$, ..., $A_n$, $B_1$, $B_2$, ..., $B_n$. Of course, since $\sum_{i=1}^{n}A_{i}+\sum_{i=1}^{n}B_{i}=\sum_{i=1}^{n}\left(A_{i}+B_{i}\right)$, this becomes $\sum_{i=1}^{n}A_{i} \cdot \sum_{i=1}^{n}B_{i}\geq \sum_{i=1}^{n}\frac{A_{i}B_{i}}{A_{i}+B_{i}}\cdot \sum_{i=1}^{n}\left(A_{i}+B_{i}\right) $, what is exactly your inequality. darij

My solution is to prove the case $n=2$ and then aply induction putting $a_n'=a_n+a_{n+1}$ and using case $n=2$

I see that this problem has allready been solved (in many ways ), but i'll post my solution cause i liked it. Let $S_a = \sum a_i$ , $S_b= \sum b_i$ and without loss of generallity let $S_a + S_b = 1$. Moreover let us assume that $(\sum a_i)(\sum b_i) < (\sum a_i+b_i)(\sum\frac{a_i\cdot b_i}{a_i+b_i})$. Then easily follows that $\sum \frac{(a_i)^2}{a_i +b_i} < (S_a)^2$ and $\sum \frac{(b_i)^2}{a_i +b_i} < (S_b)^2$. So by multiplying these two inequalities and apply C-S we get : $(\sum \frac{a_i\cdot b_i}{a_i+b_i})^2 \leq (\sum \frac{(a_i)^2}{a_i +b_i})(\sum \frac{(b_i)^2}{a_i +b_i})<(S_a\cdot S_b)^2$. This is clearly a contradiction. The rest follows.

Anto wrote: Then easily follows that $\sum \frac{(a_i)^2}{a_i +b_i} < (S_a)^2$ and $\sum \frac{(b_i)^2}{a_i +b_i} < (S_b)^2$ How ???

Since $ S_a\cdot S_b < \sum \frac{a_i\cdot b_i}{a_i+b_i} $ we get that $ S_a\cdot (1-S_a) < \sum \frac{a_i\cdot b_i}{a_i+b_i} $ or $ \sum (a_i - \frac{a_i\cdot b_i}{a_i+b_i}) < (S_a)^2$ e.t.c.

thanx, now I see , really nice solution

Thanks

This inequality is not so hard,it could be proved by mathematical induction for i. I tried that one and I get long but very simple solution