About minimum and maximum inequality.

Question

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.

Answer 1

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}

Answer 2

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.