If $a,b,c,d$ are chosen from the set $\{1,2,3,....,9\}$, then what is the minimum value of $\frac{a}{b}+\frac{c}{d} \ $?
I will show two workouts here....
1st workout
Let value of $a=1,b=9,c=2,d=8$. Then $\frac{a}{b}+\frac{c}{d} = \frac{13}{36}$
And thus the minimum value.
2nd workout
using AM-GM inequality
$\frac{a}{b}+\frac{c}{d} \geq 2 \sqrt{\frac{a c}{b d} }$
giving us the value $\frac{1}{ 3}$
Now one must see that $\frac{1}{3}$ lesser than $\frac{13}{36}$..
then what should be the correct answer to this problem
We have to minimise $\frac{a}{c} + \frac{b}{d} $. So we seek minimise the numerator and maximise the denominator of individual fractions.
Thus $(a,b) = (1,2)$ for minimum numerator. For $(c,d)$ we can have two cases $(8,9)$ or $(9,8)$. Check these two cases.
As shown in comment by Steven, $\frac{1}{9} + \frac{2}{8} > \frac{1}{8} + \frac{2}{9}$ by checking the difference.
Hence you have the minimum value as $\frac{1}{8} + \frac{2}{9} = \frac{25}{72}$