Prove when b > 0, d > 0, $ \frac{a}{b} < \frac{c}{d} $ follows $ \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$

Question

As the title already tells, I'm trying to prove the following claim:

When b > 0, d > 0, $ \frac{a}{b} < \frac{c}{d} $ follows $ \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$

My approach would be to split up the proof into 2 parts, with the 1st proving that $ \frac{a}{b} < \frac{a+c}{b+d} $ is true and with the 2nd proving that $ \frac{a+c}{b+d} < \frac{c}{d} $

However I'm struggling to come any further than this.

Important to mention is also that this problem requires you to only use field axioms.

Does this answer your question? Prove $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$. — An Alien, Nov 05 '21 at 03:31
The is the so-called “mediant inequality” and has been asked and answered many many times. You can find more identical questions with Approach0, e.g. https://math.stackexchange.com/q/751155/42969, https://math.stackexchange.com/q/1989104/42969, https://math.stackexchange.com/q/289549/42969. — Martin R, Nov 05 '21 at 07:37
And this is the generalization: https://math.stackexchange.com/q/704411/42969 — Martin R, Nov 05 '21 at 07:43
I suggest to get familiar with the Approach0 math search engine. All your questions so far turned out to be duplicates. See https://math.meta.stackexchange.com/a/29267/42969 for some information about Approach0. — Martin R, Nov 05 '21 at 07:49

score 2 · Accepted Answer · answered Nov 05 '21 at 03:09

2

We have $b>0, d>0, bc-ad > 0$.

Your first part is equivalent to proving that

$$b(a+c)-a(b+d)>0.$$

It should be clear from the given conditions that this holds.

answered Nov 05 '21 at 03:09

Siong Thye Goh

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Thank you for the short & precise answer! – On a mission Nov 05 '21 at 07:15
This has been asked and answered many many times. – Martin R Nov 05 '21 at 07:38

Prove when b > 0, d > 0, $ \frac{a}{b} < \frac{c}{d} $ follows $ \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$

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