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Ilhan M. Izmirli George Mason University Iizmirl2@Gmu.Edu Ilhan M. Izmirli George Mason University [email protected] In this paper our goal is to prove the well-known chain of inequalities involving the harmonic, geometric, logarithmic, identric, and arithmetic means using nothing more than basic calculus. Of course, these results are all well-known and several proofs of them and their generalizations have been given: Hardy, Littlewood, and Polya (1964) Carlson (1965) Carlson and Tobey (1968) Beckenbach and Bellman (1971) Alzer (1985a, 1985b) Our purpose here is to present a unified approach and give the proofs as corollaries of one elementary theorem. = For a sequence of numbers x {x1, x2 ,..., xn}we will let n ∑x j − AM (x , x ,..., x ) = AM (x) = j 1 1 2 n n n = = /1 n GM (x1, x2 ,..., xn ) GM (x) ∏x j j=1 and n = = HM (x1, x2 ,..., xn ) HM (x) n ∑ 1 j=1 x j to denote the well-known arithmetic , geometric , and harmonic means , also called the Pythagorean means ( PM ). The Pythagorean means have the obvious properties: 1. PM(x1,x2,...,xn)is independent of order 2. PM(x,x,...,x)=x = 3. PM(bx 1,bx 2,...,bx n) bPM(x1,x2,...,xn) 1. 2. is always a solution of a simple equation. In particular, the arithmetic mean (1, 2) of two numbers x1 and x 2 can be defined via the equation − = − AM x1 x 2 AM The harmonic mean satisfies the same relation with reciprocals, that is, it is a solution of the equation 1 1 1 1 − = − x1 HM HM x 2 The geometric mean of two numbers x1 and x 2 can be visualized as the solution of the equation x 1 = GM GM x 2 1. = ()() 2. 1 1 1, = 1 1 (1, ) 1 3. 1 1 1 2 (1 + 2 + + ) + + + ≥ 1 2 This follows because (1, 2, … , ) ≥ 1 (1, 2, … , ) (0, 2) = (1, 0) = 0 (1, 1) = 1 and for positive distinct numbers and 1 2 2 − 1 (1, 2) = 2 − 1 The following are some basic properties of the logarithmic means: Logarithmic mean can be thought of as , the mean-value of the function over = the interval . , The logarithmic mean can also be interpreted as the area under an exponential curve. Since $ " − $ !" $"%& = %& = − $ We also# have the identity# !" " , $ = ' $ %& Using this representation it #is easy to show that (, ($ = ((, $) We have the identity (), $)) = (, $) (, $) which follows easily: (), $)) ) − $) − $ + $ = * = (, $) 2( − $) − $ 2 The identric mean of two distinct positive real numbers is defined as: , ) -. - !- 1 ) . / + , ) = -/ , with + , = The slope of the secant line joining the points and on the graph of the (, ) (, ) function f(x) = x ln(x) is the natural logarithm of . +(, ) Theorem 1. Suppose is a function with a strictly increasing derivative.0 , 1 Then 2 6 − − & %& 3 4 5 − 5 7 2 − − () + 8 − for all in . 5 3 & , Let be defined by the equation 5# 9 − () 5# = Then, − 6 − − & %& 3 4 5# − 5# 7 2 − − () + 8 − is the sharpest form of the above inequality. Proof. By the Mean Value Theorem, for all in , we have 5, & , & − (5) 3 :(;) & − 5 for some between and . Assuming without loss of generality; by5 the assumption& of the theorem we have 5 3 &, & − 5 3 & − 5 :(&) Integrating both sides with respect to , we have & 6 & %& 3 − 5 + − − 5 7 6 − − & %& 7 and the inequality of the theorem follows. Let us now put < 5 = − 5 + − 6 − 5 − − & %& 7 Note that <9 5 = − 9 5 − − () Moreover, since 6 < = < = − + − & %& 7 there exists an in such that 9 . 5# (, ) < 5# = 0 Since is strictly increasing, we have : 9 9 < 5 = < 5# = 0 for 5 = 5# and 9 9 < 5 3 < 5# = 0 for 5 3 5# Thus, is a minimum of and for 5# < <(5#) > <(5) all in 5 , . Let us now assume that 0 3 3 . Let us let The condition of the Theorem & = . 1 is satisfied. " We compute /A 5# = And the inequality of the theorem becomes (, ) 3 (, ) Now, let us let . The condition of & = Theorem 1 is satisfied. " One can easily compute 5# = And the inequality of the theorem becomes (, ) 3 (, ) Now let . Again the condition of & = −& Theorem 1 is satisfied. The of the theorem can 5# be computed from the equation 5# = the logarithmic mean of a and b. The inequality of the theorem becomes (, ) 3 +(, ) Finally, let us put . Again the & = && condition of Theorem 1 is satisfied. In this case, 5# = + The identric mean of a and b. The inequality of the theorem becomes +(, ) 3 (, ) Thus, we now have for B (, ) 3 (, ) 3 (, ) 3 +(, ) 3 (, ) ® Alzer H. 1985a. Über einen Wert, der zwischen dem geometrischen und dem artihmetischen Mittel zweier Zahlen liegt. In Elemente Math., 40 (1985), p. 22-24. ® Alzer H. 1985b. Ungleichungen für 7 6. In Elemente Math., 40 (1985), p. 120-123. ,C C , ® Beckenbach E. F. and R. Bellman. 1971. Inequalities. Third Edition. Ergebnisse der Mathematik, Band 30, Berlin and New York: Springer- Verlag. ® Carlson, B. C. 1965. A hypergeometric mean value. In Proc. Am. Math. Soc. 16 (1965), p. 759-766. ® Carlson B. C. and M. D. Tobey. 1968. A property of the hypergeometric mean value. In Proc. Am. Math. Soc. 19 (1968), p. 255-262..
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