prove $\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$

Question

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.

Answer 1

Hint:

We only need to prove this is true for $n=2$. Since if it's ok for $n=2$, then

\begin{align} \frac{ab}{a+b} &= \frac{(a_1 + \sum_{i=2}^na_i)(b_1 + \sum_{i=2}^nb_i) }{a_1 + \sum_{i=2}^na_i + b_1 + \sum_{i=2}^nb_i} \\ &\geq \frac{a_1b_1}{a_1 + b_1} + \frac{(\sum_{i=2}^na_i)(\sum_{i=2}^nb_i)}{\sum_{i=2}^na_i + \sum_{i=2}^nb_i} \\ & \geq \cdots \geq \sum_{i=1}^n\frac{a_ib_i}{a_i + b_i} \end{align}

When $n=2$, the inequality is equivalent to \begin{align} \frac{a_1b_1}{a_1 + b_1}(a_2 + b_2) + \frac{a_2b_2}{a_2 + b_2}(a_1 + b_1) \leq a_1b_2 + a_2b_1 \end{align}

which can be proven by noting $x_1 = \frac{a_1}{a_1 + b_1}(a_2 + b_2)$, $y_2 = b_1$, $x_2 = a_2$, $y_1 = \frac{b_2}{a_2 + b_2}(a_1 + b_1)$ and using rearrangement inequality

Answer 2

This is really just a brute-force variation on the answer by Liu Gang. Consider the equivalent form $$ \sum_{i,j}\frac{a_ib_i(a_j+b_j)}{a_i+b_i}\leq\sum_{i,j}a_ib_j. $$ In these two sums, the diagonal terms with $i=j$ cancel, so we can equivalently show $$ \sum_{i\neq j}\frac{a_ib_i(a_j+b_j)}{a_i+b_i}\leq\sum_{i\neq j}a_ib_j. $$ In order to show this inequality, it is sufficient to show, for all $i\neq j$, that $$ \frac{a_ib_i(a_j+b_j)}{a_i+b_i}+\frac{a_jb_j(a_i+b_i)}{a_j+b_j}\leq a_ib_j+a_jb_i. $$ In other words, we have reduced to the case $n=2$. Let us, for notational convenience, put $i=1,j=2$. Here we could employ the rearrangement inequality to finish the argument. Let us instead multiply by $(a_1+b_1)(a_2+b_2)$ and cancel terms putting this inequality in the equivalent form $$ a_1b_1a_2b_2+a_2b_2a_1b_1\leq a_1^2b_2^2+a_2^2b_1^2. $$ Again, we could finish the argument by using the rearrangement inequality. Let us instead add $a_1^2b_1^2+a_2^2b_2^2$ to both sides, yeilding $$ \sum_{i,j=1}^2 a_i b_i a_j b_j\leq \sum_{i,j=1}^2 a_i^2b_j^2. $$ But this is the Cauchy-Schwarz inequality applied to the vectors $a=(a_1,a_2),b=(b_1,b_2)$ (which is, of course, a consequence of the rearrangement inequality).