How to prove $\sum_{i=1}^n \frac{a_i b_i}{a_i+b_i} \sum^n_{i=1} (a_i+b_i) \le (\sum_{i=1}^n a_i)(\sum_{i=1}^n b_i) ,a_i>0,b_i>0$?

$$\sum_{i=1}^n \frac{a_i b_i}{a_i+b_i} \sum^n_{i=1} (a_i+b_i) \le \left(\sum_{i=1}^n a_i\right)\left(\sum_{i=1}^n b_i\right) ,a_i>0,b_i>0$$

I have tried to use the 1st MI, but failed with so many sums. How to prove this?

We have $$\frac{a_ib_i}{a_i + b_i} = \frac{(a_i + b_i)^2 - (a_i - b_i)^2}{4(a_i + b_i)} = \frac{a_i + b_i}{4} - \frac{(a_i - b_i)^2}{4(a_i + b_i)}.$$ The desired inequality is written as $$\sum_{i=1}^n \frac{a_i + b_i}{4}\, \sum_{i=1}^n (a_i + b_i) - \sum_{i=1}^n a_i \,\sum_{i=1}^n b_i \le \frac14\sum_{i=1}^n \frac{(a_i - b_i)^2}{a_i + b_i} \,\sum_{i=1}^n (a_i + b_i)$$ or $$\frac14\left(\sum_{i=1}^n a_i - \sum_{i=1}^n b_i\right)^2 \le \frac14\sum_{i=1}^n \frac{(a_i - b_i)^2}{a_i + b_i}\, \sum_{i=1}^n (a_i + b_i).$$ Using Cauchy-Bunyakovsky-Schwarz inequality, we have $$\sum_{i=1}^n \frac{(a_i - b_i)^2}{a_i + b_i}\, \sum_{i=1}^n (a_i + b_i) \ge \left(\sum_{i=1}^n |a_i - b_i|\right)^2.$$ It suffices to prove that $$\frac14\left(\sum_{i=1}^n (a_i - b_i)\right)^2 \le \frac14\left(\sum_{i=1}^n |a_i - b_i|\right)^2$$ which is true (using triangular inequality).

We are done.

Claim: $ \frac{a_1b_1}{a_1+b_1} \times a_2b_2 + \frac{a_2b_2}{a_2+b_2} \times a_1b_1 \leq a_1b_2 + a_2b_1$

Proof: The inequality is equivalent to $ \frac{(a_1b_2-a_2b_1)^2}{(a_1+b_1)(a_2+b_2)} \geq 0 $.
Equality holds iff $ \frac{a_1}{b_1}=\frac{a_2}{b_2} $.

Corollary: The desired inequality holds.
Apply the claim to all of the cross terms.
The non-cross terms cancel directly since $ \frac{a_ib_i}{a_i+b_i} \times {a_i+b_i} = a_ib_i$.
Equality holds iff $ \frac{a_i}{b_i} $ is a constant.

Note: From this, we see that the original inequality can be written as the SOS $ \sum_{i\neq j} \frac{a_ib_j -a_jb_i}{(a_i+b_i)(a_j+b_j)} \geq 0 $, but of course that was hard to guess directly.