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?


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

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

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



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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