For positive ints a,b,c,d. Prove that if a/b < c/d, then a/b < (a+c)/(b+d) < c/d

Question

I have no idea where to start for this proof. I attempted to write that if the proposition is true, then b > d and a < c.

However, this is not necessarily true. Since for a=1, b=6, c=3, d=b the proposition is true and for a=2, b=9, c=a, d=3 the proposition is also true.

But if I write a <= c and b >= d, when a=c and b=d the proposition does not hold.

Maybe I am going at it from the wrong angle, but I can't figure out how to prove it if a relationship for a,b,c,d is not found.

https://math.stackexchange.com/q/205654/42969 , https://math.stackexchange.com/q/1989104/42969 — Martin R, Aug 10 '21 at 07:03
Hint: $\frac{a}{b}=x\leftrightarrow a=bx$ and $\frac{c}{d}=y\leftrightarrow c=dy$ — acat3, Aug 10 '21 at 07:03
Does this answer your question? Given 4 integers, $a, b, c, d > 0$, does $\frac{a}{b}<\frac{c}{d}$ imply $\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$? — Martin R, Aug 10 '21 at 07:05
More identical or similar questions with answers: https://math.stackexchange.com/questions/linked/205654. — Martin R, Aug 10 '21 at 07:21

score 0 · Accepted Answer · answered Aug 10 '21 at 07:04

0

Given $\frac{a}{b} < \frac{c}{d}$ and $a,b,c,d$ positive (positivity is important, otherwise you would have to work out the signs properly):

$\frac{a}{b} < \frac{a+c}{b+d} \iff a(b+d) < b(a+c) \iff {\color{blue}{ab}} + ad < {\color{blue}{ab}} + bc \iff ad < bc \iff \frac{a}{b} < \frac{c}{d}$.

The second inequality can be proven in a similar way.

answered Aug 10 '21 at 07:04

dictatemetokcus

585

Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Aug 10 '21 at 07:56

For positive ints a,b,c,d. Prove that if a/b < c/d, then a/b < (a+c)/(b+d) < c/d

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