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

1 Answer
Nov 18, 2017

See explanation...

Explanation:

Suppose $a , b \ge 0$ (essentially required to even have a definition of geometric mean).

Their arithmetic mean is $\frac{1}{2} \left(a + b\right)$ and their geometric mean is $\sqrt{a b}$, both of which are non-negative.

Note that:

$0 \le {\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2} = {a}^{2} + 2 a b + {b}^{2} - 4 a b = {\left(a + b\right)}^{2} - 4 a b$

Adding $4 a b$ to both ends and transposing, we find:

${\left(a + b\right)}^{2} \ge 4 a b$

Dividing both ends by $4$ that becomes:

${\left(\frac{a + b}{2}\right)}^{2} \ge a b$

Since both sides are non-negative, and square root is monotonically increasing, we can take the square root of both sides to find:

$\frac{a + b}{2} \ge \sqrt{a b}$