SOLUTION: If a < b, show that a < (a + b)/2 < b. Note: The number (a + b)/2 is called the arithmetic mean of a and b. Can someone get me started by providing the first-two steps?

Question 1199346: If a < b, show that a < (a + b)/2 < b.
Note: The number (a + b)/2 is called the arithmetic mean of a and b.
Can someone get me started by providing the first-two steps?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52818) (Show Source): You can put this solution on YOUR website!
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If a < b, show that a < (a + b)/2 < b.
Note: The number (a + b)/2 is called the arithmetic mean of a and b.
Can someone get me started by providing the first-two steps?
~~~~~~~~~~~~~~~~~

Write this equality/(identity), which is true for any/each value of "a" a = + . (1) In this equality, replace the second addend by the greater value . Doing this way, from equality (1) you will get then INEQUALITY a < + . (2) It is the same as a < . (3) Thus, first part of the statement is proved. +---------------------------------------------------+ | From this point, I continue for the second part. | +---------------------------------------------------+ Write this equality/(identity), which is true for any/each value of "b" + = b. (4) In this equality, replace first addend by the lesser value . You will get then INEQUALITY + < b. (5) It is the same as < b. (6) Combining (3) and (6) together in one compound inequality, we get a < < b, QED.

Solved and explained.

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

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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