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

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Answer 1 · 2017-11-18T13:59:03

See 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)#