By DAVID ALEXANDER BRANNAN

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An ð a1 a2 . . an Þ n : n Proof (2) Since the ai are positive, we can rewrite (2) in the equivalent form 1 ð a1 a2 . . an Þ n ða1 þ a2 þ Á Á Á þan Þ=n 1: (3) Now, replacing each term ai by lai for any non-zero number l does not alter the left-hand side of the inequality (3). It follows that it is sufficient to prove the inequality (2) in the special case when the product of the terms ai is 1. Hence it is sufficient to prove the following statement P(n) for each natural number n: P(n): For any positive real numbers ai with a1a2 .

1, and prove that P(k þ 1) is then true. Now, if all the terms a1, a2, . , akþ1 are equal to 1, the result P(k þ 1) certainly holds. Otherwise, at least two of the terms differ from 1, say a1 and a2, such that a1 > 1 and a2 < 1. Hence ða1 À 1Þ Â ða2 À 1Þ 0; which after some manipulation we may rewrite as a1 þ a 2 ! 1 þ a1 a2 : (4) We denote the typical term by ai rather than ak to avoid confusion with a different use of the letter k in the Mathematical Induction argument below. We will prove this by Mathematical Induction.

If anþ1 À an ! 0; for n ¼ 1; 2; . ;  If anþ1 À an for n ¼ 1; 2; . ; 0; then fan g is increasing. then fan g is decreasing. If an > 0 for all n, it may be more convenient to use the following version of the strategy: Strategy To show that a given sequence of positive terms, {an}, is monotonic, consider the expression aanþ1 . n   If aanþ1 ! 1; n anþ1 If an 1; for n ¼ 1; 2; . ; for n ¼ 1; 2; . ; then fan g is increasing. then fan g is decreasing. ; n ¼ 1; 2; . ; (b) an ¼ 2Àn ; n ¼ 1; 2; . ; (c) an ¼ n þ 1n ; n ¼ 1; 2; .

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