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Prove that for any rearrangement of $b_1, b_2, \cdots, b_n$ of the positive numbers $a_1,a_2,\cdots,a_n$ one has $\sum\limits_{i=1}^n \dfrac{a_i }{b_i} \geq n$.

Attempt:

First, one can write $\sum \frac{a_i}{b_i} = \sum \dfrac{a_i + b_i }{b_i} - n \geq n$ so we can prove

$$ \sum \dfrac{ a_i + b_i }{b_i} \geq 2n \iff \sum \dfrac{a_i + b_i }{2b_i} \geq n $$

and we a sum of $n$ numbers of the form $\dfrac{ a_i + b_i }{2b_i} \geq \dfrac{\sqrt{a_i b_i } }{ b_i} = \sqrt{ \dfrac{ a_i}{b_i} }$. I see a no-end from here since it is not always true that $a_i \geq b_i$.

....

Other strategy is perhaps use induction? If $n=2$, then we have $(a_1,a_2)$ an if $b_i = a_i$ then we have the result easily. If $b_1 = a_2$, then

$$ \dfrac{a_1}{a_2} + \dfrac{a_2}{a_1} \geq 2 $$

which follows because $x + 1/x \geq 2 $. Now, suppose result is true for $n$ then

$$ \sum^{n} \dfrac{a_i}{b_i} + \dfrac{a_{n+1}}{b_{n+1}} \geq n + \dfrac{a_{n+1}}{b_{n+1} } $$

how can one prove that $a_{n+1} \geq b_{n+1}$ ?

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    $\begingroup$ Maybe just a direct result of the rearrangement inequality? $\endgroup$
    – Yuta
    Commented May 8, 2020 at 3:34
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    $\begingroup$ Just apply the AM-GM inequality to the fractions $\frac{a_k}{b_k}$. $\endgroup$ Commented May 8, 2020 at 3:40
  • $\begingroup$ Got it! because $\sum \dfrac{a_i}{b_i} \geq n ( \prod \frac{a_i}{b+i} )^{1/n} = n $. Correct? Thanks profesor! Write it as an answer to reward you for your reply! $\endgroup$
    – James
    Commented May 8, 2020 at 5:06

1 Answer 1

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This would follow from the rearrangement inequality.

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