A basic logarithmic inequality, and the logarithmic mean

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1 Notes on Number Theory nd Discrete Mthemtics ISSN Vol. 2, 205, No., 3 35 A bsic logrithmic inequlity, nd the logrithmic men József Sándor Deprtment of Mthemtics, Bbeş-Bolyi University Str. Koglnicenu nr., Cluj-Npoc, Romni e-mil: jjsndor@hotmil.com Abstrct: By using the bsic logrithmic inequlity ln x x we deduce integrl inequlities, which prticulrly imply the inequlities G < L < A for the geometric, logrithmic, resp. rithmetic mens. Keywords: Logritmic function, Logrithmic men, Mens nd their inequlities. AMS Clssifiction: 26D5, 26D99. Introduction Let, b > 0. The logrithmic men L = L(, b) of nd b is defined by L = L(, b) = b for b nd L(, ) =. () ln b ln Let G = G(, b) = b nd A = A(, b) = + b denote the clssicl geometric, resp. 2 logrithmic mens of nd b. One of the most importnt inequlities for the logrithmic men (besides e.g. < L(, b) < b for < b) is the following: G < L < A for b (2) The left side of (2) ws discovered by B. C. Crlson in 966 ([] see [2]), while the right side in 957 by B. Ostle nd H. L. Terwilliger [3]. We note tht reltion (2) hs pplictions in mny subject of pure or pplied mthemtics nd physics including e.g. electrosttics, probbility nd sttistics, etc. (see e.g. [4, 5]). The following bsic logrithmic inequlity is well-known: 3

2 Theorem. ln x x for ll x > 0. (3) There is equlity only for x =. Inequlity (3) my be proved e.g. by considering the uxiliry function f(x) = x ln x, nd it is esy to show tht x = is globl minimum to f, so f(x) f() = 0. Another proof is bsed on the Tylor expnsion of the exponentil function, yielding e t = + t + t2 2 eθ, where θ (0, t). Put t = x, nd (3) follows. The continuous rithmetic, geometric nd hrmonic mens of positive, integrble function f : [, b] R re defined by nd A f = b f(x)dx, G f = e b ln f(x)dx H f = b dx/f(x) where < b re rel numbers. By using (3) we will prove the following clssicl fct:, Theorem 2. H f G f A f (4) Then, by pplying (4) for certin prticulr functions, we will deduce (2). In fct, (2) will be obtined in stronger form. The min ide of this note is the use of very simple inequlity (3) in the theory of mens. 2 The proofs Proof of Theorem 2. Put x = (b )f(t) in (3), nd integrte on t [, b] the obtined inequlity. One gets ln (( b )) (b ) (b ) (b )= 0. 32

3 This gives the right side of (4). Apply now this inequlity to f in plce of f. As ln f(t) we immeditely obtin the left side of (4). Corollry. If f is s bove, then ( This follows by H f A f in (4). ) ( = ln f(t), ) f(t) dt (b ) 2. (5) Remrk. Let f be continuous in [, b]. The bove proof shows tht there is equlity e.g. in right side of (4) if f(t) =. (6) b By the first men vlue theorem of integrls, there exists c [, b] such tht b = f(c). Since by (6) one hs f(t) = f(c) for ll t [, b], f is constnt function. When f is integrble, s ln f(t) (b ) dt = 0, (b )f(t) s for g(t) = ln > 0 one hs g(t)dt = 0, it follows by known result tht g(t) = 0 lmost everywhere (.e.). Therefore.e., thus f is constnt.e. f(t) = b Remrk 2. If f is continuous, it follows in the sme mnner, tht in the left side of (4) there is equlity only for f = constnt. The sme is true for inequlity (5). 33

4 Proof of (2). Apply G f A f to f(x) =. Remrk tht x where < I(, b) < b. b ln xdx = ln I(, b), This men is known in the literture s identric men (see e.g. [4]). As f(x) = is not x constnt, we get by A f = L(, b), G f = I(, b), tht Applying the sme inequlity G f A f to f(x) = x one obtins L < I (7) I < A (8) Remrk 3. Inequlities (7) nd (8) cn be deduced t once by pplying ll reltions of (4) to f(x) = x. Apply now (5) to f(t) = e t. After elementry computtions, we get e b e b > e +b 2 (9) As f(t) > 0 for ny t R, inequlity (9) holds true for ny, b R, b >. Replce now b := ln b, := ln, where now the new vlues of nd b re > 0. One gets from (9): By tking into ccount of (7) (0), we cn write: i.e. (2) is proved (in improved form on the right side). L > G (0) G < L < I < A, () Remrk 4. Inequlity (4) (thus, reltion (0)) follows lso by G f A f pplied to f(t) = e t. Remrk 5. The right side of (2) follows lso from (5) by the ppliction f(t) = t. As the reltion follows. tdt = b2 2 2 nd dt = (ln b ln ), t Remrk 6. Clerly, in the sme mnner s (4), the discrete inequlity of mens cn be proved, nx i by letting x = (x,..., x n > 0). x x n 34

5 References [] Crlson, B. C. (966) Some inequlities for hypergeometric functions, Proc. Amer. Mth. Soc., 7, [2] Crlson, B. C. (972) The logrithmic men, Amer. Mth. Monthly, 79, [3] Ostle, B., & Terwilliger, H. L. (957) A comprison of two mens, Proc. Montn Acd. Sci., 7, [4] Sándor, J. (990) On the identric nd logrithmic mens, Aeq. Mth., 40, [5] Lorenzen, G. (994) Why mens in two rguments re specil, Elem. Mth., 49,

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