Question 1199346
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a < (a+b)/2 < b
breaks into
a < (a+b)/2 and (a+b)/2 < b


Let's do a bit of algebraic operations on the first inequality we've broken down
a < (a+b)/2
a < 0.5(a+b)
a < 0.5a+0.5b
a-0.5a < 0.5b
0.5a < 0.5b
a < b


Reverse those steps:
a < b
0.5a < 0.5b
0.5a+0.5a < 0.5b+0.5a
a < 0.5a+0.5b
a < 0.5(a+b)
a < (a+b)/2


We can also say the following:
a < b
0.5a < 0.5b
0.5a+0.5b < 0.5b+0.5b
0.5(a+b) < b
(a+b)/2 < b



With a < b as our starting point for the last two subsections, we've found that
a < (a+b)/2 and (a+b)/2 < b
which combines into
a < (a+b)/2 < b


Therefore, the claim has been proven.



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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


Now 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 as a last set of steps:
2a < c < 2b
2a < a+b < 2b
2a/2 < (a+b)/2 < 2b/2
a < (a+b)/2 < b
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