SOLUTION: If 0 < a < b, show that a < sqrt{a•b} < b. The number sqrt{a•b} is called the geometric mean of a and b.

Question 1207707: If 0 < a < b, show that a < sqrt{a•b} < b. The number sqrt{a•b} is called the geometric mean of a and b.

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

a < sqrt(ab) < b is the ultimate goal we want to reach.

Break it into these two key parts
a < sqrt(ab) and sqrt(ab) < b

Let's focus on the first part
a < sqrt(ab)
a^2 < ab ......... square both sides
a < b ........... divide both sides by 'a'

The inequality sign won't flip when dividing both sides by 'a' since a > 0.

Let's reverse the flow of that logic to get this:
a < b
a*a < b*a
a^2 < ab
sqrt( a^2 ) < sqrt(ab)
a < sqrt(ab)

Follow a similar set of steps for the other key part.
a < b
a*b < b*b
ab < b^2
sqrt(ab) < sqrt(b^2)
sqrt(ab) < b

We have determined
a < sqrt(ab) and sqrt(ab) < b
which glue together to get
a < sqrt(ab) < b

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