Prove inequality using Positivity Axioms

Question

Given an ordered field and $b,d>0$ and $\frac{a}{b}<\frac{c}{d}$, prove that $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$ using positivity axioms.

Unfortunately, I'm stuck with this one. I feel like I've been trying to manipulate it algebraically for hours now to get some insight as to how to start the proof, but I'm not making any progress. Any hints would be greatly appreciated!

This is a duplicate of https://math.stackexchange.com/questions/1989104/mediant-inqeuality-proof — max_zorn, Jan 27 '18 at 06:10
Possible duplicate of Mediant Inequality Proof: $\frac{a}{b} < \frac{a+c}{b+d} <\frac{c}{d}$ — Micah, Jan 27 '18 at 21:28

score 1 · Answer 1 · answered Jan 27 '18 at 05:31

1

Hint: $\require{cancel}\;a(\cancel{b}+d)-b(\cancel{a}+c)=ad - bc\,$.

answered Jan 27 '18 at 05:31

dxiv

76,497

score 1 · Accepted Answer · answered Jan 27 '18 at 05:34

Assuming $b,d > 0$, we have,

$$\begin{eqnarray}\frac{a}{b} < \frac{c}{d} &\Longrightarrow& ad < bc\\ &\Longrightarrow& ad + ab < bc + ab\\ &\Longrightarrow& a(b + d) < b(a + c)\\ &\Longrightarrow& \frac{a}{b} < \frac{a+c}{b+d}\\ \end{eqnarray}$$

Try to do something similar for the other inequality.

Prove inequality using Positivity Axioms

2 Answers2