Arithmetic Mean-Geometric Mean Inequality

In both algebra and geometry, one will encounter the fact that for any positive real numbers $a$ and $b$, $\frac{a+b}{2} \ge \sqrt{ab}$, or in words, the arithmetic mean (average) is greater than or equal to the geometric mean. This can be generalized to $n$ numbers in the following way.

$$\frac{a_1+a_2+a_3+\cdots +a_n}{n}\ge (a_1a_2a_3\cdots a_n)^{\frac{1}{n}}$$

with equality for $a_1=a_2=a_3=\cdots =a_n.$ See problem #23 in the notes to Ted Alper's talk in November on Mathematical Induction [3]. See the notes to Bjorn Poonen's talk on Inequalities [4] for a much more extensive examination of the topic of inequalities with a great set of problems to work on. I didn't learn much about mathematics in the classes that I took for three and a half years to get my degree in mathematics. I found out a great deal about mathematics later, from a series of books called the New Mathematical Library which is now being published by the Mathematical Association of America. The first book in the series is Numbers: Rational and Irrational by Ivan Niven. The fifteenth book in the series is Mathematics of Choice: How to Count without Counting also by Ivan Niven. For a gentle but thorough introduction to the subject of inequalities I recommend two more volumes in the series; Introduction to Inequalities [5] and Geometric Inequalities [6]. For this talk, the case of $n=3$ will be sufficient, so we will look at an idea from Paul Zeitz's book on problem solving [7]. As every student from China, whom I have received into my classes seems to know when they arrive

$$a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$

From this we can now deduce the case of $n=3$ of the AM-GM inequality.