Chen, Chao-Ping

Qi, Feng

English

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On Integral Version of Alzer's Inequality and Martins' Inequality

Victoria University Eprints Repository

ON INTEGRAL VERSION OF ALZER’S INEQUALITY ANDMARTINS’ INEQUALITYCHAO-PING CHEN AND FENG QIAbstract. Let c > b > a and r be real numbers, and let f be a positive, twicedifferentiable function and satisfy f ′(t) > 0 and (ln f(t))′′ ≥ 0 on (a,+∞).Thensupx∈[a,b] f(x)supx∈[a,c] f(x)< 1b−a ∫ ba fr(x) dx1c−a∫ ca fr(x) dx1/r < 1 for all real r, 1b−a ∫ ba fr(x) dx1c−a∫ ca fr(x) dx1/r ≶ exp(1b−a∫ ba ln f(x) dx)exp(1c−a∫ ca ln f(x) dx) according as r ≷ 0.This solves a recently open problem of B.-N. Guo and F. Qi.1. IntroductionIt was shown in [1, 2, 8, 13, 17] that let n be a positive integer, then for r > 0,nn+ 1<(1nn∑i=1ir/1n+ 1n+1∑i=1ir)1/r<n√n!n+1√(n+ 1)!(1)We call the left-hand side of (1) H. Alzer’s inequality [1], and the right-hand sideof (1) J. S. Martins’ inequality [8]. In [3, 14] Alzer’s inequality is extended to allreal r. In [5] it was proved that Martins’ inequality is reversed for r < 0.F. Qi and B.-N. Guo [10, 11] presented an integral version of inequality (1) asfollows: Let b > a > 0 and δ > 0, then for r > 0,bb+ δ<(1b−a∫ baxr dx1b+δ−a∫ b+δaxr dx)1/r<[bb/aa]1/(b−a)[(b+ δ)b+δ/aa]1/(b+δ−a). (2)We note that the inequality (4) can be written for r > 0 asbb+ δ<Lr(a, b)Lr(a, b+ δ)<I(a, b)I(a, b+ δ), (3)where Lr(a, b) and I(a, b) are respectively the generalized logarithmic mean and theexponential mean of two positive numbers a, b, defined in [6, 15, 16] by, for a = bby Lr(a, b) = a and for a 6= b byLr(a, b) =(br+1 − ar+1(r + 1)(b− a))1/r, r 6= −1, 0;2000 Mathematics Subject Classification. Primary: 26D10, Secondary: 26D15.Key words and phrases. Alzer’s inequality, Martins’ inequality, mean.The authors were supported in part by SF of Henan Innovation Talents at Universities, China.12 CH.-P. CHEN AND F. QIL−1(a, b) =b− aln b− ln a = L(a, b);L0(a, b) =1e(bbaa)1/(b−a)= I(a, b).L(a, b) is the logarithmic mean of two positive numbers a, b. When a 6= b, Lr(a, b)is a strictly increasing function of r. In particular,limr→−∞Lr(a, b) = min{a, b}, limr→+∞Lr(a, b) = max{a, b}.In [4], it was indirectly shown that the function r 7→ Lr(a, b)/Lr(a, b+ δ) is strictlydecreasing with r ∈ (−∞,+∞). This yields thatbb+ δ<(1b−a∫ baxr dx1b+δ−a∫ b+δaxr dx)1/rfor all real r, (4)(1b−a∫ baxr dx1b+δ−a∫ b+δaxr dx)1/r≶ [bb/aa]1/(b−a)[(b+ δ)b+δ/aa]1/(b+δ−a)according as r ≷ 0. (5)In [7], B.-N. Guo and F. Qi ask under which conditions the inequalitysupx∈[a,b] f(x)supx∈[a,b+δ] f(x)<(1b−a∫ bafr(x) dx1b+δ−a∫ b+δafr(x) dx)1/r<exp(1b−a∫ baln f(x) dx)exp(1b+δ−a∫ b+δaln f(x) dx)(6)holds for b > a > 0, δ > 0 and r > 0.V. Mascioni [9] found the sufficient conditions on the function f , and proved theright-hand inequality of (6) for r > 0. Motivated by the paper of Mascioni [9], weestablish the followingTheorem. Let c > b > a and r be real numbers, and let f be a positive, twicedifferentiable function and satisfy f ′(t) > 0 and (ln f(t))′′ ≥ 0 on (a,+∞). Thensupx∈[a,b] f(x)supx∈[a,c] f(x)<(1b−a∫ bafr(x) dx1c−a∫ cafr(x) dx)1/r< 1 for all real r, (7)(1b−a∫ bafr(x) dx1c−a∫ cafr(x) dx)1/r≶exp(1b−a∫ baln f(x) dx)exp(1c−a∫ caln f(x) dx) according as r ≷ 0. (8)Both bounds in (7) are best possible.2. LemmasLemma 1. Let the function f be a positive and twice differentiable on (a,+∞),where a is a given real number, and letG(t) =1t−a∫ taf(x) dxf(t), t > a.Then we have(i) If f ′(t) > 0 and (ln f(t))′′ ≥ 0, then the function G is strictly decreasing on(a,+∞).ON INTEGRAL VERSION OF ALZER’S INEQUALITY AND MARTINS’ INEQUALITY 3(ii) If f ′(t) < 0 and (ln f(t))′′ ≤ 0, then the function G is strictly increasing on(a,+∞).Proof. Easy calculation reveals that[(t− a)f(t)]2G′(t)f(t) + (t− a)f ′(t) =(t− a)f2(t)f(t) + (t− a)f ′(t) −∫ taf(x) , H(t),[f(t) + (t− a)f ′(t)]2H ′(t)(t− a)f3(t) = −(t− a)f ′′(t)f(t)− [f ′(t)]2f2(t)− f′(t)f(t)= −[(t− a)(ln(f(t))′′ + (ln f(t))′].If (ln f(t))′ > (<)0 and (ln(f(t))′′ ≥ (≤)0 for t > a, then H ′(t) < (>)0 for t >a, and then, H(t) < (>)H(a) = 0 and G′(t) < (>)0 for t > a. The proof iscomplete. Lemma 2 ([12]). If F(t) is a strictly increasing (decreasing) integrable functionon an interval I ⊆ R, then the arithmetic mean G(r, s) of function F(t),G(r, s) =1s− r∫ srF(t) dt, r 6= s,F(r), r = s,is also strictly increasing (decreasing) with both r and s on I.3. Proof of TheoremFor r = 0, (7) can be interpreted asf(b)f(c)<exp(1b−a∫ baln f(x) dx)exp(1c−a∫ caln f(x) dx) < 1. (9)Define for t > a,P (t) =exp(1t−a∫ taln f(x) dx)f(t).A simple computation yields(t− a)2P′(t)P (t)= (t− a) ln f(t)−∫ taln f(x) dx− (t− a)(ln f(t))′ , Q(t),Q′(t) = −(t− a) [(ln f(t))′ + (t− a)(ln(f(t))′′] < 0.Hence, we have Q(t) < Q(a) = 0 and P ′(t) < 0 for t > a. This means the left-handinequality of (9) holds for c > b > a. By Lemma 2, the right-hand inequality of (9)holds clearly.For r 6= 0, (7) is equivalent tofr(b)fr(c)≶1b−a∫ bafr(x) dx1c−a∫ cafr(x) dx≶ 1, according as r ≷ 0. (10)Define for t > a,Gr(t) =1t−a∫ tafr(x) dxfr(t).4 CH.-P. CHEN AND F. QIIt is easy to see that(ln fr(t))′t ≷ 0 and (ln(fr(t))′′t T 0, according as r ≷ 0. (11)By Lemma 1, the function t 7→ Gr(t) strictly decreasesincreases with respect to t ∈(a,+∞) according as r ≷ 0. This produces the left-hand inequality of (10). ByLemma 2, the right-hand inequality of (10) holds clearly.Both bounds in (7) are best possible because oflimr→+∞(1b−a∫ bafr(x) dx1c−a∫ cafr(x) dx)1/r=supx∈[a,b] f(x)supx∈[a,c] f(x),limr→−∞(1b−a∫ bafr(x) dx1c−a∫ cafr(x) dx)1/r=infx∈[a,b] f(x)infx∈[a,c] f(x)= 1.The inequality (8) is equivalent to1b−a∫ bafr(x) dx1c−a∫ cafr(x) dx<exp(1b−a∫ baln fr(x) dx)exp(1c−a∫ caln fr(x) dx) for r 6= 0. (12)Define for t > a,Fr(t) =1t−a∫ tafr(x) dxexp(1t−a∫ taln fr(x) dx) .It is easy to see from the proof of Theorem 1 of [9] that if f ′(t) > 0 and (ln f(t))′′ ≥0, then the function F1 is strictly increasing on (a,+∞); If f ′(t) < 0 and (ln f(t))′′ ≤0, then the function F1 is strictly decreasing on (a,+∞). Applying this result,together with (11), we obviously imply the function t 7→ Fr(t) strictly increasesdecreaseswith respect to t ∈ (a,+∞) according as r ≷ 0. This produces (12). The proof iscomplete.References[1] H. Alzer, On an inequality of H. Minc and L. Sathre, J. Math. Anal. Appl. 179 (1993),396–402.[2] Ch.-P. Chen and F. Qi, Notes on proofs of Alzer’s inequality, O. Math. Mag. 11 (2003), no.1, 29–33.[3] Ch.-P. Chen and F. Qi, The inequality of Alzer for negative powers, O. Math. Mag. 11 (2003),no. 2, 442–445.[4] Ch.-P. Chen and F. Qi,Monotonicity properties for generalized logarithmic means, AustralianJournal of Mathematical Analysis and Applications 1 (2004), no.2, Article 2. Available onlineat http://ajmaa.org/[5] Ch.-P. Chen, F. Qi and S. S. Dragomir, Reverse of Martins’ inequality, Australian Journalof Mathematical Analysis and Applications, 2 (2005), no.1, Article 2. Available online athttp://ajmaa.org/[6] L. Galvani, Dei limiti a cui tendono alcune media, Boll. Un. Mat. Ital. 6 (1927), 173–179.[7] B.-N. Guo and F. Qi, An algebraic inequality, II, RGMIA Res. Rep. Coll. 4 (2001), no.1,Article 8, 55–61. Available online at http://rgmia.vu.edu.au/v4n1.html.[8] J. S. Martins, Arithmetic and geometric means, an application to Lorentz sequence spaces,Math. Nachr. 139 (1988), 281–288.ON INTEGRAL VERSION OF ALZER’S INEQUALITY AND MARTINS’ INEQUALITY 5[9] V. Mascioni, A sufficient condition for the integral version of Martins’ inequality, J. Inequal.Pure Appl. Math. 5 (2004), no. 2, Artitle 32. Available online at http://jipam.vu.edu.au/issues.php?op=viewissue&issue=78.[10] F. Qi, An algebraic inequality, J. Inequal. Pure Appl. Math. 2 (2001), no. 1, Artitle 13.Available online at http://jipam.vu.edu.au/v2n1/006_00.html.[11] F. Qi and B.-N. Guo, An inequality between ratio of the extended logarithmic means andratio of the exponential means, Taiwanese J. Math. 7 (2003), no. 2, 229-237.[12] F. Qi, Logarithmic convexity of extended mean values, Proc. Amer. Math. Soc. 130(2002), no. 6, 1787–1796. Available online at http://www.ams.org/journal-getitem?pii=S0002-9939-01-06275-X.[13] J. Sa´ndor, On an inequality of Alze, J. Math. Anal. Appl. 192 (1995), 1034–1035.[14] J. Sa´ndor, On an inequality of Alzer, II, O. Math. Mag. 11 (2003), no. 2, 554–555.[15] K. B. Stolarsky, Generalizations of the logarithmic mean, Math. Mag. 48 (1975), 87–92.[16] K. B. Stolarsky, The power and generalized logarithmic means, Amer. Math. Monthly, 87(1980), 545–548.[17] J. S. Ume, An elementary proof of H. Alzer’s inequality, Math. Japon. 44 (1996), no. 3,521–522.(Ch.-P. Chen) Department of Applied Mathematics and Informatics, Research In-stitute of Applied Mathematics, Henan Polytechnic University, Jiaozuo City, Henan454010, ChinaE-mail address: chenchaoping@hpu.edu.cn, chenchaoping@sohu.com(F. Qi) Department of Applied Mathematics and Informatics, Research Institute ofApplied Mathematics, Henan Polytechnic University, Jiaozuo City, Henan 454010, ChinaE-mail address: qifeng@hpu.edu.cn, fengqi618@member.ams.orgURL: http://rgmia.vu.edu.au/qi.html, http://dami.hpu.edu.cn/qifeng.html

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On Integral Version of Alzer's Inequality and Martins' Inequality

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