Question 1207706
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We want to show that a < (a + b)/2 < b


This breaks down into these two key parts
a < (a + b)/2  and  (a + b)/2 < b


What I'll do is start with the first piece and follow these steps
a < (a + b)/2
2a < a + b
2a-a < a + b - a
a < b


Next, I'll reverse those steps by following the chain upward.
a < b
a+a < b+a
2a < a+b
a < (a+b)/2



We have gone from the given a < b to prove that a < (a+b)/2 is the case.


Follow similar steps for the other needed key component.
a < b 
a+b < b+b 
a+b < 2b 
(a+b)/2 < b


We have shown that if a < b, then 
a < (a + b)/2  and  (a + b)/2 < b
in which those two pieces glue together to get
a < (a + b)/2 < b


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Another approach


Let c = a+b


Add 'a' to both sides of the original inequality.
a < b
a+a < b+a
2a < a+b
2a < c


Return back to the original inequality.
This time add b to both sides.
a < b
a+b < b+b
c < 2b


Since 2a < c and c < 2b, we know that 2a < c < 2b.


Then,
2a < c < 2b
2a < a+b < 2b
2a/2 < (a+b)/2 < 2b/2
a < (a+b)/2 < b
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