How to prove an inequality

Question

$a$, $b$, $c$, $d$ are rational numbers and all $> 0$.

$\max \left\{\dfrac{a}{b} , \dfrac{c}{d}\right\} \geq \dfrac{a+c}{b+d}\geq \min \left\{\dfrac{a}{b} , \dfrac{c}{d}\right\}$

Hope someone can help me with this one. How would you go on proving the validity? Thanks in advance.

Answer 1

$$\frac{a+c}{b+d}-\frac ab=\frac{b(a+c)-a(b+d)}{(b+d)b}=\frac{bc-ad}{b(b+d)}$$

Similarly, $$\frac{a+c}{b+d}-\frac cd=\cdots=\frac{ad-bc}{(b+d)d}$$

Observe that the signs of the terms are opposite as $a,b,c,d>0$

Answer 2

Let $m=\min\{a/b,c/d\}$. This means that $$ m \le \frac ab \qquad \text{and} \qquad m\le \frac cd,$$ which is equivalent to $bm\le a$ and $dm\le c$. Thus we get $$a+c\ge bm+dm=(b+d)m$$ which is equivalent to $$\frac{a+c}{b+d}\ge m.$$

The proof of the inequality for maximum is similar.

(Note that in all steps, where we multiplied the inequality by some number, this number was positive. Therefore the sing of the inequality was not changed.)