An Application of the Am-Gm Inequality

From my archives back in 2008... I preceded this with an explanation of both means to lay a foundation for the students tying the algebraic and geometric relations together.


The arithmetic mean-geometric mean (Am-Gm) inequality is equal in only one situation, when all the numbers are the same. It would seem such an obvious statement would be of almost no mathematical use, but in fact, it is a very useful tool. If you graph y=(x)^1/3 you notice that the function is always increasing. That means if a^1/3 > b^1/3 then a>b. That leads to a rather big idea in algebra/geometry.
When we seperate a number into parts that add up to that number it is called a partition of the number. One partition of 12 into three values is 6 + 4 + 2. What we get from the Am-Gm inequality is that the products of any partition is greatest when the elements of the partition are all equal. For partitions of 12 into three numbers, the highest product is 4(4)(4)=64.
Try any of others, it will always be less. (6)(4)(2) is only 48; and 6(3)(3)= 54. There are others, but none of the partitions of three numbers will be as large as 64.
You may be intuitively familiar with an example of this from geometry. Take any rectangle with a given perimeter; the one with the largest area is a square. In fact, the square is the largest area for any quadrilateral with a given perimeter.
You can prove the same propetry for triangles algebraicly with Heron's formula. You remember Heron's formula; for any trinangle with sides a, b, and c, we find the semi-perimeter S= (a+b+c)/2. Then the area is given by [S (s-a)(s-b)(s-c)]^1/2. If we hold the perimeter constant, then s^1/2 is a constant, and we can rewrite the area as the product of two square roots, (s)^1/2 times [(s-a)(s-b)(s-c)]^1/2. But since (s)^1/2 is a constant, the area is larger or smaller depending on the product of (s-a)(s-b)(s-c); and from the Am-Gm inequality we know that this product is largest when all three values are equal. But s-a = s-b=s-c means that a=b=c, so the maximum area for any perimeter is when the triangle is equilateral.
With a little geometric effort, you can show that the same is true for a polygon of any number of sides, the one with the greatest area for any perimeter is the regular polygon.

Concave polygons have an angle that juts into the interior of the shape, and convex ones do not. It should be easy to see that if a shape was concave you could just flip the concave part out and keeping all the sides exactly the same length you have created
a shape with a larger area, so trivially (always be suspicious when that word is used, even now) a maximal area for any perimeter will be convex. Now imagine a convex polygon and focus on three consecutive vertices.

Now create an ellipse that has its two foci (that is the plural of focus) at the two extreme vertices, and passes through the third point. One of the properties you ought to remember about an ellipse is that the sum of the distances from any point on the ellipse to the two foci is a constant, so if we move the one point along the ellipse, we will not be changing the sum of t he lengths of the two sides, and the perimeter of the polygon is staying constant.
Now what is happening to the interior area of the polygon as we move this point along the ellipse. The only part of the interior of the polygon that changes is inside the triangle formed by the three vertices. But this triangle has a constant base, and as we move the third point around the ellipse.... we are changing the height of the triangle. The area is the greatest where the height is the greatest, and that happens when the altitude rises from the midpoint of the two foci; and at that point, the triangle is isosceles. The area is greatest if we make these two sides of the polygon equal. Now pick three other points, and repeat endlessly until all the exterior sides are congruent to maximize the area.