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How can i prove the following inequality

$$\ \min{\left(\frac{a_1}{b_1},\frac{a_2}{b_2}\right)}\leq\frac{a_1+a_2}{b_1+b_2}\leq\max{\left(\frac{a_1}{b_1},\frac{a_2}{b_2}\right)} $$ for any real numbers $a_1,a_2$ and positive numbers $b_1,b_2$.

Can anyone give me some hint or reference for the proof of this inequality?

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marked as duplicate by Martin R, Arnaud D., Guy Fsone, user223391, Community Nov 28 '17 at 15:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ (These questions add the conditions that the number are integers, but the answers do not use this condition.) $\endgroup$ – Arnaud D. Nov 28 '17 at 13:46
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WLOG, let $\frac{a_1}{b_1}\le \frac{a_2}{b_2}$

$\implies \min{\left(\frac{a_1}{b_1}, \frac{a_2}{b_2}\right)} = \frac{a_1}{b_1}$,

$\implies \max\left(\frac{a_1}{b_1},\frac{a_2}{b_2}\right) = \frac{a_2}{b_2}$

Now, use the fact $a_1 \le \frac{a_2b_1}{b_2}$ in $\frac{a_1+a_2}{b_1+b_2}$ to get

$\frac{a_1+a_2}{b_1+b_2} \le \frac{a_2}{b_2}$

Similarly, use the fact $a_2 \ge \frac{a_1b_2}{b_1}$ in $\frac{a_1+a_2}{b_1+b_2}$ to get

$\frac{a_1+a_2}{b_1+b_2} \ge \frac{a_1}{b_1}$

QED

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Suppose without loss of generalities that $$\frac{a_1}{b_1}\geq \frac{a_2}{b_2}.\tag{1}$$ Now assume by contradiction that $$\frac{a_1+a_2}{b_1+b_2}>\frac{a_1}{b_1}=\max\big\{\frac{a_1}{b_1}, \frac{a_2}{b_2}\big\},$$ then we have $$(a_1+a_2)b_1 >a_1(b_1+b_2) \qquad \implies \qquad a_2b_1>a_1b_2\qquad \implies \qquad \frac{a_2}{b_2}>\frac{a_1}{b_1},$$ a contradiction to $(1)$.

The inequality for the $\min$ can be proved in the same way.

For a reference, you can check the proof of Lemma 4.3 in this paper.

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