If $a_i$ and $b_i$ are positive, and $b= \sum ^n_{i=1} b_i$,$a= \sum ^n_{i=1} a_i$ prove $$\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$$ Additional: we should just use Cauchy inequality. However , if it's not possible, solve it with anything you want.
Things I have done so far:
Writing in different form $\frac {(\sum ^n_{i=1} b_i)(\sum ^n_{i=1} a_i)}{(\sum ^n_{i=1} b_i)+(\sum ^n_{i=1} a_i)} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$
And maybe this could take me somewhere: ${(\sum ^n_{i=1} b_i)(\sum ^n_{i=1} a_i)} \geq (\sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i})({(\sum ^n_{i=1} b_i)+(\sum ^n_{i=1} a_i)})$
And another thing which came to my mind:
${(\sum ^n_{i=1} b_i)(\sum ^n_{i=1} a_i)} \geq (\sum ^n_{i=1} \sqrt {a_ib_i})^2$
$(\sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i})(\sum ^n_{i=1} a_i+b_i)\geq (\sum ^n_{i=1} \sqrt {a_ib_i})^2$
My main problem is find two parentheses to multiply them for writing Cauchy. I want some hint on this just to start working with inequality.
Update
It seems like no one can come with answer using Cauchy and without induction and Harmonic , Geometric and ... means.so if no answer come till tomorrow,I will give up and accept Liu Gang answer.
And I would appreciate someone could explain why it can't be solved using Cauchy and without induction and Harmonic , Geometric and ... means.(asking this because possibility of putting bounty)
2nd-Update
we can use any mean inequalities.but for only two number like $x$ and $y$.not generalized form for $n$ numbers.