Geometric Mean

[greatwhiteshark]

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

 

[Denis]

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

d = b - a
a = b - d [1]
b = a + d) [2]

[1] sqrt(ab) = sqrt((b-d)b) = sqrt(b^2 - bd) < b

[2] sqrt(ab) = sqrt(a(a+d)) = sqrt(a^2 + ad) > a

 

[pka]

Here is another way for this one.

 

[greatwhiteshark]

Thank You

I can see where this leads to.