How do you prove that arithmetic mean is greater than equal to geometric mean?
1 Answer
Nov 18, 2017
See explanation...
Explanation:
Suppose
Their arithmetic mean is
Note that:
0 <= (a-b)^2 = a^2-2ab+b^2 = a^2+2ab+b^2-4ab = (a+b)^2-4ab
Adding
(a+b)^2 >= 4ab
Dividing both ends by
((a+b)/2)^2 >= ab
Since both sides are non-negative, and square root is monotonically increasing, we can take the square root of both sides to find:
(a+b)/2 >= sqrt(ab)