A question that remains is whether this procedure can be adjusted to work for neg- ative values of r. Since I don’t know the answer, I will leave it to interested readers.

Acknowledgments. I wish to thank Professors Fred Gass and Tom Farmer for their helpful comments.

References

1. Elizabeth M. Markham, Geometric series, Mathematics Magazine 66 (1993) p. 242. 2. Roger B. Nelsen, Proofs Without Words: Exercises in visual thinking, Classroom Resourse Materials, The Mathematical Association of America, Washington, 1993. 3. , Proofs Without Words II: More Exercises in Visual Thinking, Classroom Resourse Materials, The Mathematical Association of America, Washington, 2000. 4. Warren Page, Geometric sums, Mathematics Magazine 54 (1981) p. 201.

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The HM-GM-AM-QM Inequalities Philip Wagala Gwanyama ([email protected]), Northeastern Illinois University, Chicago, IL 60625

Many sources have discussed one or more of the inequalities involving harmonic mean, geometric mean, arithmetic mean, and quadratic mean (see [1], [2], [3], [5], [6]). Kung [4] provided a geometric proof without words of the inequalities connecting the harmonic mean, geometric mean, arithmetic mean and quadratic mean (or root mean square) for two variables. In this note, we use the method of Lagrange multipli- ers, to discuss the inequalities for more than two variables. For positive real numbers x1, x2,...,xn, we show that 1/n n n n 2 n = x j = x j < ≤ ≤ j 1 ≤ j 1 . 0 n 1 x j = = n n j 1 x j j 1

The harmonic mean–geometric mean–arithmetic mean inequalities. To prove these inequalities, we let

n n g(x1, x2,...,xn) = x j , s(x1, x2,...,xn) = x j , j=1 j=1 n ( , ,..., ) = h x1 x2 xn n 1 = j 1 x j and find bounds for g subject to the constraint s = A,whereA is a constant. We illustrate the method for n = 3 by proving that

3 x + x + x ≤ (x x x )1/3 ≤ 1 2 3 . 1 + 1 + 1 1 2 3 3 x1 x2 x3

VOL. 35, NO. 1, JANUARY 2004 THE COLLEGE MATHEMATICS JOURNAL 47 To do so, we solve for x1, x2, x3,andλ in the system of equations −→ −→ ∇ g(x1, x2, x3) = λ ∇ s(x1, x2, x3) and s(x1, x2, x3) = A. +  + = λ + λ + λ 3 = From x2x3ı x1x3 x1x2k ı k and j=1 x j A, we obtain the only = = = A critical point x1 x2 x3 3 , which gives g’s maximum value A A A A3 g , , = 3 3 3 27

A3 for the constraint S(x1, x2, x3) = A. To see this, observe that g < for any other + 27 point in the interior of the triangular plate {(x1, x2, x3) ∈ |x1 + x2 + x3 = A}. Each point on the boundaries of the plate has at least one x j = 0 (for j = 1, 2, 3), and so g = 0 on the boundary. Thus,

/ 3 1 3 3 j=1 x j 0 < x j ≤ . (1) j=1 3

By substituting the positive real numbers x2x3, x1x3, x1x2 into (1), we obtain x x + x x + x x (x x .x x .x x )1/3 ≤ 2 3 1 3 1 2 2 3 1 3 1 2 3 and hence,

3x1x2x3 1/3 ≤ (x1x2x3) . x2x3 + x1x3 + x1x2 Therefore, 3 3x x x x + x + x < = 1 2 3 ≤ ( )1/3 ≤ 1 2 3 . 0 1 1 1 x1x2x3 (2) + + x2x3 + x1x3 + x1x2 3 x1 x2 x3 For n > 3, the same approach can be used.

The geometric mean–arithmetic mean–quadratic mean inequalities. To prove these inequalities, we let

n n s(x1, x2,...,xn) = x j , g(x1, x2,...,xn) = x j , j=1 j=1 n 2 r(x1, x2,...,xn) = x j , j=1 and find bounds for s subject to the constraints g = Po and r = Qo,wherePo and Qo are constants. Using the method of Lagrange multipliers for n = 3, we show that x + x + x x 2 + x 2 + x 2 (x x x )1/3 ≤ 1 2 3 ≤ 1 2 3 . 1 2 3 3 3

To do so, we solve for x1, x2, x3,λo and µo in the system of equations:

48 c THE MATHEMATICAL ASSOCIATION OF AMERICA  −→ −→ −→  ∇ ( + + ) = λ ∇ ( ) + µ ∇ 2 + 2 + 2  x1 x2 x3 o x1x2x3 o x1 x2 x3 (3) x x x = P (4)  1 2 3 o  2 2 2 x1 + x2 + x3 = Qo (5) where the constants and λo and µo are the Lagrange multipliers. Equation (3) gives

ı +  + k = (λox2x3 + 2µox1) ı + (λox1x3 + 2µox2)  + (λox1x2 + 2µox3) k or

λox2x3 + 2µox1 = λox1x3 + 2µox2 = λox1x2 + 2µox3 = 1 which yields x = x = x . Using (4) and (5), we obtain the critical points x = x = √ 1 2 3 1 2 = 3 = = = Qo ( , , ) x3 Po and x1 x2 x3 3 . Thus, the bounds of s x1 x2 x3 are √ 3 3 3 3 3 s Po, Po, Po = 3 Po = 3 x1x2x3 and Q Q Q Q x 2 + x 2 + x 2 s o , o , o = 3 o = 3 1 2 3 . 3 3 3 3 3 √ 2+ 2+ 2 3 x1 x2 x3 Applying (1), we find that 3 x1x2x3 must be a lower bound of s and 3 3 an upper bound of s. Therefore, √ x + x + x x 2 + x 2 + x 2 3 x x x ≤ 1 2 3 ≤ 1 2 3 . (6) 1 2 3 3 3

Combining (2) and (6) we get, 1/3 3 3 3 2 3 = x j = x j 0 < ≤ x ≤ j 1 ≤ j 1 . 3 1 j = = 3 3 j 1 x j j 1

For n > 3, the same approach can be used.

Acknowledgments. I am grateful to the referees and previous editor Warren Page for their very useful comments and suggestions.

References

1. H. Alzer, A proof of the arithmetic mean-geometric mean inequality, The American Mathematical Monthly 103 (1996) 585. 2. P. S. Bullen, D. S. Mitrinovic, and P. M. Vasic, Means and Their Inequalities, (Dordrecht, Holland: D. Reidel Publishing Company), 1988. 3. F. Dubeau, Weighted means of order r and related inequalities: An elementary approach, The College Mathe- matics Journal 23 (1992) 211–213.

VOL. 35, NO. 1, JANUARY 2004 THE COLLEGE MATHEMATICS JOURNAL 49 4. S. Kung, Harmonic, geometric, arithmetic, root mean inequality, The College Mathematics Journal, 21 (1990) 227. 5. G. H. Hardy, J. E. Littlewood, and G. Polya,´ Inequalities (2nd ed.), Cambridge University Press, Cambridge, 1952. 6. N. Schaumberger, The AM-GM inequality via x1/x , The College Mathematics Journal 20 (1989) 320.

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Cauchy’s Mean Value Theorem Involving n Functions Jingcheng Tong ([email protected]), University of North Florida, Jacksonville, FL 32224

If f (x) and g(x) are two functions continuous on [a, b] and differentiable on (a, b) with g(x) = 0foranyx in (a, b), then there exists a point c in (a, b) such that

f (c) f (b) − f (a) = . (1) g(c) g(b) − g(a) The above Generalized Mean Value Theorem was discovered by Cauchy ([1] or [2]), and is very important in applications. Since Cauchy’s Mean Value Theorem involves two functions, it is natural to wonder if it can be extended to three or more functions. If so, what formulas similar to (1) can we have? In this capsule we show, and then extend, the following result.

Theorem 1. Let α, β be two real numbers such that α + β = 1.Iff(x),g(x), h(x) are three functions continuous on [a, b] and differentiable on (a, b) such that g(b) = g(a) and h(b) = h(a), then there exists a point c in (a, b) such that f (b) − f (a) f (b) − f (a) f (c) = αg(c) + βh(c) . (2) g(b) − g(a) h(b) − h(a)

Observe that (2) follows by letting γ =−1 and setting f1 = f , f2 = g,and f3 = h in Theorem 2.

Theorem 2. Let α, β and γ be three real numbers such that α + β + γ = 0.Iff1, f2 and f3 are three functions continuous on [a, b] and differentiable on (a, b) such that fi (a) = fi (b) for i = 1, 2, 3, then there exists a point c in (a, b) such that γ α β ( ) + ( ) + ( ) = . f1 c f2 c f3 c 0 (3) f1(b) − f1(a) f2(b) − f2(a) f3(b) − f3(a)

Proof. Let k(x) = γ f2(b) − f2(a) f3(b) − f3(a) f1(x) − f1(a) + α f1(b) − f1(a) f3(b) − f3(a) f2(x) − f2(a) + β f1(b) − f1(a) f2(b) − f2(a) f3(x) − f3(a) .

It is easily checked that k(a) = 0and k(b) = f1(b) − f1(a) f2(b) − f2(a) f3(b) − f3(a) (α + β + γ)= 0.

50 c THE MATHEMATICAL ASSOCIATION OF AMERICA