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The general inequality is : \begin{equation}\mathrm{min}(\frac{a_1}{b_1},\dots,\frac{a_n}{b_n}) \leq \frac{a_1+\dots+a_n}{b_1+\dots+b_n} \leq \mathrm{max}(\frac{a_1}{b_1},\dots,\frac{a_n}{b_n})\end{equation}where $a_1,\dots,a_n,b_1,\dots,b_n$ are real non-zero positive numbers. I thought of starting by the easier case $n=2$ but I still have no idea how to compare $\frac{a}{b}$ and $\frac{a'}{b'}$.

Ansper
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1 Answers1

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Idea: Go case by case

Case 1:

$\mathrm{min}(\frac{a}{b},\frac{a'}{b'}) = \frac{a}{b}$ and $\mathrm{max}(\frac{a}{b},\frac{a'}{b'}) = \frac{a}{b}$

Case 2:

$\mathrm{min}(\frac{a}{b},\frac{a'}{b'}) = \frac{a}{b}$ and $\mathrm{max}(\frac{a}{b},\frac{a'}{b'}) = \frac{a'}{b'}$

Case 3:

$\mathrm{min}(\frac{a}{b},\frac{a'}{b'}) = \frac{a'}{b'}$ and $\mathrm{max}(\frac{a}{b},\frac{a'}{b'}) = \frac{a}{b}$

Case 4:

$\mathrm{min}(\frac{a}{b},\frac{a'}{b'}) = \frac{a'}{b'}$ and $\mathrm{max}(\frac{a}{b},\frac{a'}{b'}) = \frac{a'}{b'}$

Noureddine Ouertani
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  • Note that the same question has been asked and answered many many times before. In such a case it is preferable to flag it as a duplicate (see https://math.stackexchange.com/help/privileges/flag-posts for more information about flagging). – Martin R Nov 01 '20 at 11:31