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?
 A: 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.
A: 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.
A: Your Inequality is equivalent to;
$\sum\frac{a_ib_i}{a_i+b_i}\le\frac{\sum a_i\sum b_i}{\sum a_i+b_i}\tag1$
I'll make use of the following inequality:
$$ \frac{1}{\sum_{i=1}^{n}\frac{1}{x_i+y_i}} \geq \frac{1}{\sum_{i=1}^{n}\frac{1}{x_i}}+\frac{1}{\sum_{i=1}^{n}\frac{1}{y_i}} \tag2$$
Proof of $(2)$: normalize both $x_i$ and $y_i$, i.e. $x_i:=\frac{x_i}{x_i+y_i}$ and $y_i:=\frac{y_i}{x_i+y_i}$
then its enough to show that,
$\frac{1}{n}\ge \frac{1}{\sum_i\frac{x_i+y_i}{x_i}}+\frac{1}{\sum_i\frac{x_i+y_i}{y_i}}\tag3$
each sum on the RHS can be seen as a Harmonic mean with missing factor $n$, apply HM$\le$ AM to the first term on the right
$\frac{1}{\sum_i\frac{x_i+y_i}{x_i}}=\frac{1}{\sum_i\frac{1}{\frac{x_i}{x_i+y_i}}}\le \frac{\sum_i\frac{x_i}{x_i+y_i}}{n^2}\tag4$
Same result is valid for the second term as well;
$\frac{1}{\sum_i\frac{x_i+y_i}{y_i}}=\frac{1}{\sum_i\frac{1}{\frac{y_i}{x_i+y_i}}}\le \frac{\sum_i\frac{y_i}{x_i+y_i}}{n^2}\tag5$
adding both I get,
$$\frac{\sum_i\frac{x_i}{x_i+y_i}}{n^2}+\frac{\sum_i\frac{y_i}{x_i+y_i}}{n^2}=\frac1n$$
So $(3)$ is indeed true.
Now change variables in $(2)$, $x_i=\frac1{a_i}$ and $y_i=\frac1{b_i}$,
$$ \frac{1}{\sum_{i=1}^{n}\frac{1}{\frac{1}{a_i}+\frac{1}{b_i}}} \geq \frac{1}{\sum_{i=1}^{n}\frac{1}{\frac{1}{a_i}}}+\frac{1}{\sum_{i=1}^{n}\frac{1}{\frac{1}{b_i}}} = \frac{1}{\sum_{i=1}^{n}a_i }+ \frac{1}{\sum_{i=1}^{n} b_i}$$
$$\frac{1}{\sum_{i=1}^{n} \frac{a_i+b_i}{a_i\cdot b_i}}\ge\frac{1}{\sum_{i=1}^{n}a_i }+ \frac{1}{\sum_{i=1}^{n} b_i}$$
$$\frac{1}{\sum_{i=1}^{n} \frac{a_i+b_i}{a_i\cdot b_i}}\ge \frac{{\sum_{i=1}^{n} a_i}+{\sum_{i=1}^{n} b_i}}{{\sum_{i=1}^{n} a_i}\cdot {\sum_{i=1}^{n} b_i}}$$
invert both sides and you get $(1)$
