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I am having trouble proving the following:

if $a < b$, then $a < {a+b\over2} < b$.

I started with the Trichotomy Property and getting to where $a^2>0$, but then I do not know where to go from there.

Any suggestions?

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Hint: $2a=a+a<a+b<b+b=2b$.

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  • $\begingroup$ All in one line. $\endgroup$ – IAmNoOne Jun 12 '14 at 20:56
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HINT: It holds that $\frac{a + a}{2} = a$. If $a<b$, what do you know about $\frac{a +a}{2} + \frac{b-a}{2}$? Can you make a similar argument for the second inequality?

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Start with $a<b$, add $a$ to both sides then time both sides by $\frac{1}2$. You will get $a< \frac{a+b}2$.

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In particular, we have the following inequality chain:

$$\min(a,b) \le \frac{2}{\frac{1}a+\frac{1}b} \le \sqrt{ab} \le \frac{a+b}2 \le \sqrt{\frac{a^2+b^2}2} \le \max(a,b).$$

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$a<b \iff \frac{a}{2}<\frac{b}{2} \iff a-\frac{a}{2}<\frac{b}{2} \iff a<\frac{a}{2}+\frac{b}{2}=\frac{a+b}{2}$

Similary $\frac{a+b}{2}<b$.

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Seeing as $(a+b)/2$ is the avg of $a$ & $b$...

The $a \lt b$ gives you a nice ordering, telling you $a$ is closest to $-\infty$. Thereby, the avg of $a$ & $b$ will be between $a$ & $b$ as opposed to being between $b$ & $a$.

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  • $\begingroup$ This is certainly intuitively appealing, but far from a rigurous answer. $\endgroup$ – vonbrand Jun 12 '14 at 21:12
  • $\begingroup$ Rigor is the easy part...seeing what to make rigorous is the tough part. $\endgroup$ – Hoble Jun 12 '14 at 21:58
  • $\begingroup$ I can't seem to get the formatting for this site correctly, but anyways...proving an avg is between two numbers is fairly simple and there are a lot of ways to do it. Simply take the case where a = b. It is then equal to the left and to the right. By making a < b, you thereby make (a + b)/2 less than b. The same logic applies to a. $\endgroup$ – Hoble Jun 12 '14 at 22:35

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