How do you prove that arithmetic mean is greater than equal to geometric mean?

1 Answer
Nov 18, 2017

See explanation...

Explanation:

Suppose a, b >= 0 (essentially required to even have a definition of geometric mean).

Their arithmetic mean is 1/2(a+b) and their geometric mean is sqrt(ab), both of which are non-negative.

Note that:

0 <= (a-b)^2 = a^2-2ab+b^2 = a^2+2ab+b^2-4ab = (a+b)^2-4ab

Adding 4ab to both ends and transposing, we find:

(a+b)^2 >= 4ab

Dividing both ends by 4 that becomes:

((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)