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

Algebra ->  Inequalities -> SOLUTION: If 0 < a < b, show that a < sqrt{ab} < b. Note: The number sqrt{ab} is called the geometric team of a and b.       Log On


   



Question 1208896: If 0 < a < b, show that a < sqrt{ab} < b.
Note: The number sqrt{ab} is called the geometric team of a and b.

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

0 < a < b

Multiply through by a:

a%2A0+%3C+a%2Aa+%3C+a%2Ab

0+%3C+a%5E2+%3C+ab

Nonnegative square roots are in the same order of inequality as their squares. 
So we take nonnegative square roots:

0+%3C+a+%3C+sqrt%28ab%29

0 < a < b

Multiply through by b:

a%2Ab+%3C+b%2Ab

a%2Ab+%3C+b%5E2

Nonnegative square roots are in the same order of inequality as their squares. 
So we take nonnegative square roots:

sqrt%28ab%29%3Cb

0+%3C+a+%3C+sqrt%28ab%29 and sqrt%28ab%29%3Cb

Therefore matrix%281%2C2%2C0+%3C+a%3Csqrt%28ab%29%3C+b%2C%22%22%29

Edwin