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The Geometric Mean and the AM-GM Inequality

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The Geometric Mean and the AM-GM Inequality The Geometric Mean and the AM-GM Inequality John Treuer February 27, 2017 1 Introduction: The arithmetic mean of n numbers, better known as the average of n numbers is an example of a mathematical concept that comes up in everday conversation. We hear about averages during weather reports and teachers tell us the class average on the most recent test. However, there are other means besides the arithmetic mean. A less commonly known mean is the geometric mean. Tonight we will investigate the geometric mean, derive the arithmetic mean-geometric mean (AM-GM) inequality and do challenging problems. 2 Review of the Arithmetic Mean and Introduc- tion to the Geometric Mean The arithmetic mean (AM) of n numbers, better known as the average of n numbers, is the most common type of mean and it is defined by a + a + + a AM(a , ..., a )= 1 2 ··· n . 1 n n The geometric mean (GM) of n numbers is the nth root of the product of n numbers; that is, GM(a , ..., a )=Ôn a a . 1 n 1 ··· n Example 2.1 Calculate the arithmetic and geometric mean of 2, 4, and 8. 1 2 + 4 + 8 14 AM(2, 4, 8)= = = 4.6. 3 3 GM(2, 4, 8)=Ô3 2 4 8 = Ô3 64 = 4. · · Problem 2.2 Fill in the missing blanks in the following table. In the last column write down whether the arithmetic mean or the geometric mean is greater. The first row of the table has been done as an example. Arithmetic Mean Geometric Mean AM or GM (AM) (GM) bigger? (8, 8) 8 8 Equal (4, 16) (2, 32) (4, 4, 4) (2, 4, 8) (1, 9) (2, 8) (5,5) (23, 33, 53) 2 We can learn a few interesting things from the previous table. For the means we calculated, the arithmetic mean was always bigger than the geometric mean except when all of the numbers being averaged were the same; in that case, the arithmetic mean and geometric mean were the same. Now, let’s show that the latter observation holds in general: Problem 2.3 Show that if a = = a = a, then GM(a , ..., a )=AM(a , ..., a ). 1 ··· n 1 n 1 n It is also true in general that for all positive numbers a1, ..., an (not necessarily equal) that GM(a , ..., a ) AM(a , ..., a ). This inequality is called the arithmetic mean- 1 n Æ 1 n geometric mean inequality and we will prove that it is true in the next section. In the previous table, you found that GM(23,33,53)=30. It is also true that GM(22 3, 32 5, 52 2)=30 because · · · (22 3)(32 5)(52 2)=23 33 53. · · · · · Problem 2.4 Can you come up with another set of numbers (a, b, c) so that GM(a, b, c)= 30? At this point you may be wondering what is so geometric about the geometric mean? Let’s give one possible explanation. 3 Problem 2.5 Consider the following rectangles. Write down the area, A, of the rectangles and the geometric means of the side lengths: 4 A = 2 A = 16 32 GM(4, 16) = GM(2, 32) = Problem 2.6 What is the side length of the square that has the same area as the rectangles in the previous problem? Write down the side length of the square and the area of the square below. A = 4 It looks like that given a rectangle, the geometric mean of the side lengths is the side length of the square that has the same area as the rectangle. Is this true? Is this also true for 3-dimensional rectangles (called rectangular prisms) and 3-dimensional squares (called cubes)? Problem 2.7 Given the rectangular prism shown below, show that the geometric mean of the side lengths equals the side length of the cube that has the same volume as the rectangular prism. 2 2 3 Normally, we only deal with 2-dimensional or 3-dimensional rectangles because hu- mans can only see in 3-dimensions. However, we can define an n-dimensional rectan- gle as a rectangle with side lengths a1, a2, ..., an. The volume of the n-dimensional rectangle is a a a . 1 · 2 ··· n Problem 2.8 Given an n-dimensional prism with side lengths a1, ..., an show that GM(a1, ..., an) equals the side length of the n-dimensional cube with the same n- dimensional area as the prism. 5 We now have a geometric meaning for the geometric mean; the geometric mean of a1, ..., an gives the side length of the square that has the same area/volume as the rectangle with side lengths a1, ..., an. Now let’s return to our previous question of why the geometric mean of n positive numbers is less than the arithmetic mean of the same n positive numbers. 3 AM-GM inequality In this section we will do some fun problems that use the AM-GM inequality and prove a simplified version of the inequality. The AM-GM inequality is stated below: AM-GM Inequality If a1, ...an are positive numbers, then a + + a Ôn a a 1 ··· n 1 ··· n Æ n and equality holds if and only if a = = a . 1 ··· n Use the AM-GM inequality to solve the following four problems. Problem 3.1 The condition that all of the an’s be positive is important. For n odd, give an example where some of the an’s are negative and the AM-GM inequality does not hold. a b Problem 3.2 Use the AM-GM inequality to show that for a, b>0, b + a 2. a/b+b/a Ø (Hint: Apply the AM-GM inequality to 2 first). 6 Problem 3.3 If xyz = 27 and x, y, z are all positive, then what is the minimum value of x + y + z. For what values of x, y, z does this minimum occur? Problem 3.4 Let r1, ..., rn be n positive numbers and let x>0. Show that n r1 + + rn (x + r1)(x + r2) (x + rn) x + ··· . ··· Æ 3 n 4 (Hint: Take the nth root of both sides first.) The AM-GM inequality allows us to do cool problems like the ones you just did. Now let’s investigate some proofs of the AM-GM inequality. When n = 2, we can give a geometric proof of the AM-GM inequality. For the next two problems you need to remember the Pythagorean theorem: If a, b, and c are the side lengths of a right triangle and c is the length of the hypotenuse of the triangle, then a2 + b2 = c2. Consider the two circles below with diameters a and b with a>bshown on the next page: 7 CLASSROOM CAPSULES Edited by Warren Page Classroom Capsules serves to convey new insights on familiar topics and to enhance pedagogy through shared teaching experiences Its format consists primarily of readily understood mathematics capsules which make their impact quickly and effectively Such tidbits should be nurtured, cultivated, and presented for the benefit of your colleagues elsewhere Queries, when available, will round out the column and serve to open further dialog on specific items of reader concern Readers are invited to submit material for consideration to: Warren Page Department of Mathematics New York City Technical College 300 Jay Street Brooklyn, NY Pythagorean11201 Theorem: b c a2 + b2 = c2 Behold! The Arithmetic-Geometric Mean Inequality Roland H. Eddy, Memorial University of Newfoundland, St. John's, Newfound? land, Canada a a + b > \fab , with equality if and only if a = b. O x B C Problem 208 3.5 Determine the length of OC, the length of BC, and OB = x in terms of a and b. Conclude that when a = b, that Ôab < a+b . ” 2 This content downloaded from 128.195.69.132 on Tue, 27 Dec 2016 06:26:48 UTC All use subject to http://about.jstor.org/terms 8 Now consider the following triangle inscribed in the semicircle with center O below: D d h c A a B O b C Problem 3.6 Let r be the radius of the semicircle. Express r in terms of a and b. Is h r or r h? Æ Æ Problem 3.7 There are three right triangles in the picture above. For each triangle, list the vertices and the corresponding Pythagorean equation in the chart below. An example has been done for you. Triangle Pythagorean Equation ABD a2 + h2 = d2 9 Problem 3.8 1) Using the three Pythagorean equations you found in the previous exercise, and the comparison between h and r in problem 3.6, solve for h in terms of a and b. Conclude that GM(a, b) AM(a, b). Æ 2)When does equality hold? (Think geometrically!) Why is it implied that this argu- ment only works for a, b 0. Ø We now have two proofs that if a and b are positive, then Ôab a+b . The next step Æ 2 is to show that for all n 2, Ø a + + a Ôn a a 1 ··· n . 1 ··· n Æ n Let’s look back at the first table you filled in at the beginning of this packet. You found that (a1, a2) GM(a1, a2) AM(a1, a2) (1, 9) 3 5 (2,8) 4 5 (5,5) 5 5 10 Notice that in the above table, as you go down the rows, a1 is increased by the same amount that a2 is decreased.
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