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Let $a_i , b_i \geqslant 0$ $\forall i \in \{1 , 2 , 3 , .... , n\}$

Prove that $$min\{\frac{a_i}{b_i}\} \leqslant \frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i} \leqslant max\{\frac{a_i}{b_i}\}$$

Given that $min\{...\}$ and $max\{...\}$ equals to the minimum and the maximum element in the set.

I don’t know how to deal with minimum and maximum. I would like some hints (or solution) please. Thank you.

Community
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Oleg Farenskiy
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2 Answers2

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Here is yet another proof, which is really elementary.

We begin by observing that \begin{align} \sum_{i=1}^n a_i = \sum_{i=1}^n \frac{a_i}{b_i}b_i \end{align} From this identity we have \begin{align} \sum_{i=1}^n a_i = \sum_{i=1}^n \frac{a_i}{b_i}b_i \geq \min\left\{\frac{a_i}{b_i}\right\}\sum_{i=1}^n b_i \end{align} and \begin{align} \sum_{i=1}^n a_i = \sum_{i=1}^n \frac{a_i}{b_i}b_i \leq \max\left\{\frac{a_i}{b_i}\right\}\sum_{i=1}^n b_i \end{align} Putting these two together we obtain \begin{align} \min\left\{\frac{a_i}{b_i}\right\}\leq \frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n b_i} \leq \max\left\{\frac{a_i}{b_i}\right\} \end{align}

Mdoc
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    This is kinda nitpicky, but we can't outright divide by $b_i$. To start with we only have $b_i \geq 0$ for all $i \in {1, \cdots, n}$, so at some point where $b_i = 0$ a term in that sum could be undefined – Matheus Andrade May 31 '20 at 05:56
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    @MatheusAndrade: The original question itself makes no sense if some $b_i$ is zero. – Martin R May 31 '20 at 06:00
  • @MartinR you're right, I hadn't thought of that. Thanks for pointing it out! – Matheus Andrade May 31 '20 at 06:15
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Denote the maximum of $\frac {a_i} {b_i}$ by $c$. Then $a_i \leq cb_i$ for all $i$. Sum this over $i$ and divide by $\sum b_i$ to get the right hand inequality. The other inequality is proved similarly.

Kavi Rama Murthy
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