A beautiful inequality, the arithmetic -mean-geometric-mean inequality (AM-GM) seems reasonable but hard to prove. It states that the usual average of n positive real numbers is at least as big as the n-th root of the product. We’ll experiment with it, guess and believe a few things about it, and then actually use our confidence in those guesses to fashion a remarkably simple proof. Here’s an example:
The average of the three numbers 4, 5, 6 is clearly 5. A consequence of AM-GM is that the cube root of the product (4)(5)(6)=120 must be less than 5. Since the cube root of 120 is less than the 5 (i.e. the cube root of 125), we gain some confidence that it works. As a natural-born-devil’s-advocate, we try to change the example so as to give the geometric mean a better chance. Keeping the average at 5, the best choice for maximizing the geometric mean seems to be: 5, 5, 5. Thus, all the numbers are equal. In this extreme case, both the arithmetic and the geometric means must be the same. In fact this suggests a proof.
Keep the arithmetic average the same and make the geometric average as big as possible. We suspect that the only way to achieve this is to make all the numbers equal. Let’s just believe this and see what happens. Notice that “keeping the average at 5” really means keeping the sum at 15; also “maximizing the geometric mean” actually suggests maximizing the product. When we replaced “4, 5, 6” with “5, 5, 5” we really needed to compare the product (4)(6) with (5)(5). This is a consequence of the difference of squares formula
Difference of Squares: (x — a)(x + a) = x^2 — a^2
with x=5 and a=1. In this example we effectively only replace two numbers—the 4 and the 6—with their average—5 with another 5.
So let’s try a more challenging example—three numbers 4, 4 and 7. If we try the same trick, perhaps we replace the 4 and 7 with a couple of 5.5’s. This gets us a bigger product, just as in the above example, but we don’t seem to be getting closer to proving that having all 5’s yields the maximum product.
Here’s where the belief in the answer helps us: Instead of moving the 4 and the 7 to their average, we move the 4 to the average (namely 5) and move the 7 the same distance in the other direction (to 6).
There are many ways to see why the product (5)(6) should be bigger than the original (4)(7), but we defer to the difference of squares formula once again. This time let x=5.5 and compare what we get with a=1.5 versus a=0.5 to see exactly what’s happening. (Don’t skip over this step—it’s the heart of the matter).
Now we have three numbers 5, 4 and 6 with the same sum, but a bigger product. This triple should look familiar (our first example!)—we remember that the next step is to replace the 4 and 6 with a couple of 5’s.
The experienced mathematician now instantaneously fills in all the details of a proof, arguing that if there is any number other than the arithmetic average, then there must be one both less than the average and another one more than average. So if any of the numbers are other than average we can make the product bigger, and bigger and bigger…etc…
However the main idea can be understood (perhaps at a different level) by elementary school students, provided they are comfortable with the difference of squares formula.
Toward that end, let’s run through a little game on five numbers 8, 8, 9, 9, 10. The total is 44 and the average is 8.8. We repeatedly replace this list of five numbers with another list, always with total 44, and at each stage the product is bigger. Also at each stage, at least one of the numbers is replaced with the average. Thus:
8, 8, 9, 9, 10—>8.8, 8, 9, 9, 9.2
We replaced the first 8 and the last 10. Now the product is bigger—why? What choice of x? What choices for a?
Keep going, replacing the two bold numbers by moving the closer one to the average 8.8 and moving the other one the same amount in the other direction:
8.8, 8, 9, 9, 9.2—>8.8, 8.4, 9, 9, 8.8
8.8, 8.4, 9, 9, 8.8—>8.8, 8.6, 8.8, 9, 8.8
8.8, 8.6, 8.8, 9, 8.8—>8.8, 8.8, 8.8, 8.8, 8.8
It’s just crazy! If any of the numbers are other than average, we can compare it to the situation where all the numbers are average, in a small number of steps. It’s slick, it’s fun and it works.
1. We just had our last Putnam session at Norwich this semester. We solved a problem using the AM-GM and just for fun tried to prove AM-GM. This post came out of that discussion. Thanks to Wise D. and Jessica F. for their role in this discussion! I asked Prof. Rob P. afterwards if he knew of it and he directed me towards this article:
An inductive proof of the AM-GM inequality, by Kong-Ming Chong, American Mathematical Monthly, Vol. 83, No. 5 (May, 1976), p. 369.
As is often the case with mathematical expositions, the problem is that you need to be something of an expert in mathematics to even realize that the article is saying the same thing as our exposition in this post! So we hope you enjoyed it and can do something like this with your students.
2. Originally in our Putnam group, we used calculus, instead of the difference of squares formula. We considered two positive numbers with a fixed sum of S, trying to maximize the product by using the function f(x)=x(S-x). The breakthrough came when we realized that it’s not just about maximizing f, but rather knowing the intervals of increase, since our “belief” in this post led us to increasing f rather than maximizing it. In this discussion I replace calculus with the most elementary idea possible, the difference of squares, to make it as clear, accessible and fun as possible.
3. “Fix the arithmetic average and make the geometric average as big as possible” means the same thing as “Keep the arithmetic average the same and make the geometric average as big as possible.” At the last minute I opted for the latter phrasing.