Unusual Inequality Proof

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Aug 2, 2016
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Pulled this out of an old textbook of mine. I'm betting this was easy for me twenty plus years ago but now I'm drawing a blank.

Let \(\displaystyle p_0, p_1, q_0, q_1 \in \mathbb{R}^{+}\).

Let \(\displaystyle \dfrac{p_0}{q_0} < \dfrac{p_1}{q_1} \). Explain why this implies that \(\displaystyle \dfrac{p_0}{q_0} < \dfrac{p_0 + p_1}{q_0 + q_1} < \dfrac{p_1}{q_1}\).

I'm baffled how to deal with that addition in the denominator. Can someone help kickstart my brain? I don't think I need a whole solution if someone can kindly just point me in the right direction.
 
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Multiply both sides of each inequality by one denominator, then the other.
 
I'm sorry but that was also my first thought. I really don't see how that helps. That gets me \(\displaystyle p_0 q_1 < p_1 q_0 \) ... and I'm no closer to figuring out how to deal with that addition in the denominator.
 
I'm sorry but that was also my first thought. I really don't see how that helps. That gets me \(\displaystyle p_0 q_1 < p_1 q_0 \) ... and I'm no closer to figuring out how to deal with that addition in the denominator.

Ok, \(\displaystyle p_0 q_1 < p_1 q_0 \). Now do the same thing with the first inequality that you need to prove.
 
Pulled this out of an old textbook of mine. I'm betting this was easy for me twenty plus years ago but now I'm drawing a blank.
Let \(\displaystyle p_0, p_1, q_0, q_1 \in \mathbb{R}^{+}\).
Let \(\displaystyle \dfrac{p_0}{q_0} < \dfrac{p_1}{q_1} \). Explain why this implies that \(\displaystyle \dfrac{p_0}{q_0} < \dfrac{p_0 + p_1}{q_0 + q_1} < \dfrac{p_1}{q_1}\).
I'm baffled how to deal with that addition in the denominator. Can someone help kickstart my brain? I don't think I need a whole solution if someone can kindly just point me in the right direction.
To avoid subscripts let \(\displaystyle a=p_0,~b=p_1,~x=q_0,~y=q_1\) now we have:
\(\displaystyle \begin{array}{l} \dfrac{a}{x} < \frac{b}{y}\\ ay < bx\\ ay + by < bx + by\\ (a + b)y < b(x + y)\\ \dfrac{{a + b}}{{x + y}} < \dfrac{b}{y} \end{array}\)______ this gives us ______\(\displaystyle \dfrac{p_0 + p_1}{q_0 + q_1} < \dfrac{p_1}{q_1}\)


Can you finish?
 
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