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Prove that for any positive numbers $a_i,b_i,$, $i=1,2,\ldots,n$ the following inequality holds: $$\frac{\prod_{i=1}^n(a_i+b_i)}{\sum_{i=1}^n(a_i+b_i)}\geq \frac{\prod_{i=1}^na_i}{\sum_{i=1}^na_i} + \frac{\prod_{i=1}^nb_i}{\sum_{i=1}^nb_i}.$$

Context: This is an exercise (without the solution) in the book for 15-19 years old students preparing to participate in math competitions. This is in the short chapter about the inequality between the sum of minima and the minimum of sum.

Before this exercise there was an example: Consider $f_i(x) = (a_i+b_i)x^2+2a_ix+a_i$. Then $$\min f_i = \frac{a_ib_i}{a_i+b_i}$$ and $$\min\sum f_i = \frac{\left(\sum_{i=1}^na_i\right)\cdot \left(\sum_{i=1}^nb_i\right)}{\left(\sum_{i=1}^na_i\right)+\left(\sum_{i=1}^nb_i\right)}.$$ Since the sum of minima is smaller than the minimum of the sum, we get the inequality $$ \sum_{i=1}^n \frac{a_ib_i}{a_i+b_i}\leq\frac{\left(\sum_{i=1}^na_i\right)\cdot \left(\sum_{i=1}^nb_i\right)}{\left(\sum_{i=1}^na_i\right)+\left(\sum_{i=1}^nb_i\right)}.$$ The desired inequality is somhow dual to this one, but I don't know how to prove it.

I would need a function $f$ (maybe multivariate) with coefficients $a_1,\ldots a_n$ such that $\min f = \prod a_i/\sum a_i$ and a function $g$ with coefficients $b_1,\ldots,b_n$ such that $\min g = \prod b_i/\sum b_i$. Moreover, $f+g$ should be a function with the same structure as $f$ and $g$ but with coeficients $a_i+b_i$. Then we would be done.

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    $\begingroup$ I believe that induction works (though it isn't motivating). $\endgroup$
    – Calvin Lin
    Oct 23, 2023 at 15:58

2 Answers 2

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Hint 1: Rewrite the inequality that we want to prove as $$ \frac { (\sum a_i )(\sum b_i) } { \sum (a_i + b_i) } \geq \frac{ (\prod a_i )(\sum b_i) + (\prod b_i )(\sum a_i)} { \prod (a_i + b_i)}.$$

Hint 2: Rewrite the prior exercise into a similar form, namely we have shown that:

$$ \frac { (\sum a_i )(\sum b_i) } { \sum (a_i + b_i) } \geq \frac{ \sum a_i b_i \prod_{j \neq i } (a_i + b_j ) } { \prod (a_i + b_i) }. $$


For the induction approach, repeatedly apply the base case of

$$ \frac {( a_1 + b_1)(a_2 + b_2) } { a_1 + b_1 + a_2 + b_2 } - \frac{ a_1 a_2 }{ a_1 + a_2 } - \frac{ b_1 b_2 } { b_1 + b)2 }\\ = \frac{ (a_1 b_2 - a_2 b_1 ) ^ 2 } { ( a_1 + a_2 )(b_1 + b_2 ) ( a_1 + b_1 + a_2 + b_2 ) } \geq 0 . $$

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  • $\begingroup$ Clever idea. So far I understand the first approach, but I'll try the induction approach as well. $\endgroup$
    – Mateo
    Oct 23, 2023 at 17:51
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Alternative proof.

If $n = 2$, we have $$\mathrm{LHS} - \mathrm{RHS} = \frac{(a_1 b_2 - a_2 b_1)^2}{(a_1 + a_2)(b_1 + b_2)(a_1 + a_2 + b_1 + b_2)} \ge 0.$$

In the following, assume that $n \ge 3$.

Multiplying both sides by $\sum_{i=1}^n a_i + \sum_{i=1}^n b_i$, the desired inequality is written as $$\prod_{i=1}^n (a_i + b_i) \ge \prod_{i=1}^n a_i + \prod_{i=1}^n b_i + \frac{\sum_{i=1}^n b_i}{\sum_{i=1}^n a_i}\prod_{i=1}^n a_i + \frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i}\prod_{i=1}^n b_i. \tag{1}$$

We have $$\frac{\sum_{i=1}^n b_i}{\sum_{i=1}^n a_i}\prod_{i=1}^n a_i = \sum_{k=1}^n \left(b_k \cdot \frac{a_k}{\sum_{i=1}^n a_i}\cdot \prod_{i=1, \, i \ne k}^n a_i\right) \le \sum_{k=1}^n \left(b_k\prod_{i=1, \, i \ne k}^n a_i\right).$$

Thus, it suffices to prove that $$\prod_{i=1}^n (a_i + b_i) \ge \prod_{i=1}^n a_i + \prod_{i=1}^n b_i + \sum_{k=1}^n \left(b_k\prod_{i=1, \, i \ne k}^n a_i\right) + \sum_{k=1}^n \left(a_k\prod_{i=1, \, i \ne k}^n b_i\right). \tag{2}$$

RHS of (2) has $2n + 2$ terms $$\prod_{i=1}^n a_i, \quad \prod_{i=1}^n b_i, \quad b_k\prod_{i=1, \, i \ne k}^n a_i, \quad a_k\prod_{i=1, \, i \ne k}^n b_i, \quad k = 1, 2, \cdots, n$$ which are pairwise different; Moreover, clearly each term is also in LHS of (2) after full expanding. Thus, (2) is true.

We are done.

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