# 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)#