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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!

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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.

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

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