OCTOGON MATHEMATICAL MAGAZINE Vol 17, No1, April 009, pp 1-9 ISSN 1-5657, ISBN 978-973-8855-5-0, wwwhetfaluro/octogon 1 Various proofs of the Cauchy-Schwarz inequality Hui-Hua Wu and Shanhe Wu 0 ABSTRACT In this paper twelve different proofs are given for the classical Cauchy-Schwarz inequality 1 INTRODUCTION The Cauchy-Schwarz inequality is an elementary inequality and at the same time a powerful inequality, which can be stated as follows: Theorem Let (a 1, a,,a n and (b 1, b,,b n be two sequences of real numbers, then ( ( ( a i b i, (1 with equality if and only if the sequences (a 1, a,,a n and (b 1, b,,b n are proportional, ie, there is a constant λ such that a k = λb k for each k {1,,,n} As is known to us, this classical inequality plays an important role in different branches of modern mathematics including Hilbert spaces theory, probability and statistics, classical real and complex analysis, numerical analysis, qualitative theory of differential equations and their applications (see [1-1] In this paper we show some different proofs of the Cauchy-Schwarz inequality 0 Received: 31008 000 Mathematics Subject Classification 6D15 Key words and phrases Cauchy-Schwarz inequality; arithmetic-geometric means inequality; rearrangement inequality; mathematical induction; scalar product
Hui-Hua Wu and Shanhe Wu SOME DIFFERENT PROOFS OF THE CAUCHY-SCHWARZ INEQUALITY Proof 1 Expanding out the brackets and collecting together identical terms we have j=1 (a i b j a j b i = = ( j=1 ( b j + j=1 a j ( a i b i a i b i b j a j = Because the left-hand side of the equation is a sum of the squares of real numbers it is greater than or equal to zero, thus ( ( ( a i b i j=1 Proof Consider the following quadratic polynomial f (x = ( ( x a i b i x + = (a i x b i Since f(x 0 for any x R, it follows that the discriminant of f(x is negative, ie, ( ( a i b i The inequality (1 is proved ( 0 Proof 3 When n = 0 or n We can now assume that =0, (1 is an identity A n = 0, B n = 0, x i = a i An, y i = b i Bn (i = 1,,,n,
Various proofs of the Cauchy-Schwarz inequality 3 then x i = The inequality (1 is equivalent to yi = 1 x 1 y 1 + x y + x n y n 1, that is (x 1 y 1 + x y + x n y n x 1 + x + + x n + y 1 + y + + y n, or equivalently (x 1 y 1 + (x y + + (x n y n 0, which is evidently true The desired conclusion follows Proof 4 Let A = a 1 + a + + a n, B = b 1 + b + + b n By the arithmetic-geometric means inequality, we have so that Thus a i b i AB = a i b i AB ( 1 a i A + b i B = 1, a 1 + a + + a n b 1 + b + + b n ( ( a i b i ( Proof 5 Let A n = a 1 +a + +a n, B n = a 1 b 1 +a b + +a n b n, C n = b 1 +b + +b n It follows from the arithmetic-geometric means inequality that A n C n B n + 1 = C n B n + C n = ( a i C n B n + b i C n a i b i B n =, therefore
4 Hui-Hua Wu and Shanhe Wu that is A n C n B n, (a 1 + a + + a n(b 1 + b + + b n (a 1 b 1 + a b + + a n b n Proof 6 Below, we prove the Cauchy-Schwarz inequality by mathematical induction Beginning the induction at 1, the n = 1 case is trivial Note that (a 1 b 1 + a b = a 1b 1 + a 1 b 1 a b + a b a 1b 1 + a 1b + a b 1 + a b = = (a 1 + a (b 1 + b, which implies that the inequality (1 holds for n = Assume that the inequality (1 holds for an arbitrary integer k, ie, ( k ( k a i b i ( k Using the induction hypothesis, one has k+1 k+1 = k + a k+1 k + b k+1 k k + a k+1b k+1 k k+1 a i b i + a k+1 b k+1 = a i b i It means that the inequality (1 holds for n = k + 1, we thus conclude that the inequality (1 holds for all natural numbers n This completes the proof of inequality (1 Proof 7 Let A = {a 1 b 1, a 1 b n, a b 1,, a b n,, a n b 1, a n b n } B = {a 1 b 1, a 1 b n, a b 1,, a b n,, a n b 1, a n b n }
Various proofs of the Cauchy-Schwarz inequality 5 C = {a 1 b 1, a 1 b n, a b 1,, a b n,, a n b 1, a n b n } D = {a 1 b 1, a n b 1, a 1 b,, a n b,, a 1 b n, a n b n } It is easy to observe that the set A and B are similarly sorted, while the set C and D are mixed sorted Applying the rearrangement inequality, we have (a 1 b 1 (a 1 b 1 + + (a 1 b n (a 1 b n + (a b 1 (a b 1 + + (a b n (a b n + +(a n b 1 (a n b 1 + + (a n b n (a n b n (a 1 b 1 (a 1 b 1 + + (a 1 b n (a n b 1 + +(a b 1 (a 1 b + + (a b n (a n b + + (a n b 1 (a 1 b n + + (a n b n (a n b n, which can be simplified to the inequality (a 1 + a + + a n(b 1 + b + + b n (a 1 b 1 + a b + + a n b n as desired Proof 8 By the arithmetic-geometric means inequality, one has for λ > 0, a i b i 1 ( λ + b i λ / Choosing λ = in the above inequality gives b i a a i b i a i i + b i Hence a i b i 1 a i +,
6 Hui-Hua Wu and Shanhe Wu or equivalently a i b i 1 n + n = n n The desired conclusion follows Proof 9 Construct the vectors α = (a 1, a,, a n, β = (b 1, b,, b n Then for arbitrary real numbers t, one has the following identities for scalar product: (α + tβ (α + tβ = α α+ (α βt+(β βt α + (α βt+ β t = Thus Using the expressions = α + tβ 0 (α β α β 0 α β = a 1 b 1 + a b + + a n b n, α = we obtain ( ( a i b i, β = ( 0, Proof 10 Construct the vectors α = (a 1, a,, a n, β = (b 1, b,, b n From the formula for scalar product: we deduce that Using the expressions α β = α β cos (α, β, α β α β
Various proofs of the Cauchy-Schwarz inequality 7 α β = a 1 b 1 + a b + + a n b n, α = we get the desired inequality (1, β =, Proof 11 Since the function f (x = x is convex on (, +, it follows from the Jensen s inequality that (p 1 x 1 + p x + + p n x n p 1 x 1 + p x + + p n x n, ( where x i R, p i > 0 (i = 1,,,n, p 1 + p + + p n = 1 Case I If b i 0 for i = 1,,,n, we apply x i = a i /b i and p i = /(b 1 + b + + b n to the inequality ( to obtain that that is ( a1 b 1 + a b + + a n b n b 1 + b + + a 1 + a + + a n b n b 1 + b + +, b n (a 1 b 1 + a b + + a n b n (a 1 + a + + a n(b 1 + b + + b n Case II If there exists b i1 = b i = = b ik = 0, one has ( a i b i = i i 1,,i k,1 i n i i 1,,i k,1 i n i i 1,,i k,1 i n This completes the proof of inequality (1 Proof 1 Define a sequence {S n } by a i b i ( ( S n = (a 1 b 1 + a b + + a n b n ( a 1 + a + + a n( b 1 + b + + b n Then
8 Hui-Hua Wu and Shanhe Wu S n+1 S n = (a 1 b 1 + a b + + a n+1 b n+1 ( a 1 + a + + a n+1 (b 1 + b + + b n+1 (a1 b 1 + a b + + a n b n + ( a 1 + a + + a n which can be simplified to (b 1 + b + + b n, S n+1 S n = [ = (a 1 b n+1 b 1 a n+1 + (a b n+1 b a n+1 + + (a n b n+1 b n a n+1 ], so We thus have which implies the inequality (1 S n+1 S n (n N S n S n 1 S 1 = 0, Acknowledgements The present investigation was supported, in part, by the innovative experiment project for university students from Fujian Province Education Department of China under Grant No14, and, in part, by the innovative experiment project for university students from Longyan University of China REFERENCES [1] Dragomir, S S, Discrete inequalities of the Cauchy-Bunyakovsky-Schwarz type, Nova Science Publishers, Inc, Hauppauge, NY, 004 [] Mitrinović, D S, Pečarić, J E, Fink, A M, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht, 1993 [3] Masjed-Jamei, M, Dragomir, S S, Srivastava, HM, Some generalizations of the Cauchy-Schwarz and the Cauchy-Bunyakovsky inequalities involving four free parameters and their applications, RGMIA Res Rep Coll, 11 (3 (008, Article 3, pp1 1 (electronic
Various proofs of the Cauchy-Schwarz inequality 9 [4] Barnett, N S, Dragomir, S S, An additive reverse of the Cauchy Bunyakovsky Schwarz integral inequality, Appl Math Lett, 1 (4 (008, 388 393 [5] Lee, E Y, A matrix reverse Cauchy Schwarz inequality, Linear Algeb Appl, 430 ( (009, 805 810 [6] Dragomir, S S, A survey on Cauchy Bunyakovsky Schwarz type discrete inequalities, J Inequal Pure Appl Math, 4 (3 (003, Article 63, pp1 14 (electronic [7] Dragomir, S S, On the Cauchy Buniakowsky Schwarz inequality for sequences in inner product spaces, Math Inequal Appl, 3 (000, 385 398 [8] De Rossi, A, Rodino, L, Strengthened Cauchy Schwarz inequality for biorthogonal wavelets in Sobolev spaces, J Math Anal Appl, 99 (1 (004, 49 60 [9] Liu, Z, Remark on a Refinement of the Cauchy Schwarz inequality, J Math Anal Appl, 18 (1 (1998, 13 1 [10] Alzer, H, On the Cauchy-Schwarz inequality, J Math Anal Appl, 34 (1 (1999, 6 14 [11] Alzer, H, A refinement of the Cauchy-Schwartz inequality, J Math Anal Appl, 168 ( (199, 596 604 [1] Steiger, W L, On a generalization of the Cauchy Schwarz inequality, Amer Math Monthly, 76 (1969, 815 816 Department of Mathematics and Computer Science, Longyan University, Longyan, Fujian 36401, pr China E-mail: wushanhe@yahoocomcn