Let $a_1,a_2,\dots,a_n$ and $b_1,b_2,\dots,b_n$ be two sets of reals such that $b_j>0$ for $1\leq j\leq n$. Let m and M be respectively the minimum and maximum of n fractions $$\frac{a_1}{b_1},\frac{a_2}{b_2},\dots,\frac{a_n}{b_n}$$. Prove that $$m\leq\frac{a_1+a_2+\dots+a_n}{b_1+b_2+\dots+b_n}\leq M$$.


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    $\begingroup$ I see an edit war. $\endgroup$ – Myridium Apr 16 '17 at 4:38

For each $1 \le i \le n$, $mb_i \le a_i \le Mb_i$ since $b_i > 0$. Therefore, $$m(b_1+\dots+b_n) \le a_1+\dots+a_n \le M(b_1+\dots+b_n).$$ Now divide by $\sum b_i$.

  • $\begingroup$ $a_i$ can be negative!! $\endgroup$ – Dayal Kumar Apr 16 '17 at 4:38
  • $\begingroup$ So.............. $\endgroup$ – mathworker21 Apr 16 '17 at 4:40

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