On the Iyengar Madhava Rao Nanjundiah inequality and its hyperbolic version

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1 Notes on Number Theory and Discrete Mathematics Print ISSN , Online ISSN Vol. 4, 018, No., DOI: /nntdm On the Iyengar Madhava Rao Nanjundiah inequality and its hyperbolic version József Sándor Department of Mathematics, Babeş Bolyai University Str. Kogălniceanu 1, Cluj-Napoca, Romania Received: 1 April 016 Accepted: 9 January 018 Abstract: We provide a new proof of the trigonometric inequality obtained by K. S. K. Iyengar, B. S. Madhava Rao and T. S. Nanjundiah in 1945, and offer also the hyperbolic version of this result. Certain related results are pointed out, too. Keywords: Inequalities, Trigonometric functions, Hyperbolic functions, Iyengar Madhava Rao Nanjundiah inequality, Adamović Mitrinović inequality, Lazarović inequality, l Hospital rule of monotonicity. 010 Mathematics Subject Classification: 6D05, 6D07, 6D15, 6D99. 1 Introduction Inequalities for trigonometric and hyperbolic functions have been studied etensively in the last 0 years. Among such inequalities we mention the famous inequalities by Adamović Mitrinović, which states that > cos (1.1 for any ( 0, π ; and the Lazarović inequality > cosh (1. for any > 0. For a basic reference to (1.1 and (1., see []. See also [4, 6]. 14

2 In a little known paper [1] Iyengar, Madhava Rao and Nanjundiah have proved that ( > cos > cos (1. for ( In a recent paper [7], we have discovered (1. by an application of the famous Hadamard integral inequality. By using the same method, we have proved also the hyperbolic analogue of (1., namely ( > cosh > cosh (1.4 for any > 0. We note that, the second inequality of (1.4 follows by a comparison of certain series inequalities. Iyengar, Madhava Rao and Nanjundiah have considered by other arguments, the signs of the functions cos µ a( and showed that for µ one has a( > 0, and for µ π arccos π we have a( < 0, therefore, < cos µ, (1.5 ( 0, π is also true. In fact, µ1 and µ above are the best constants such that these inequalities are valid. The aim of this paper is to give a new proof of the above results, and also a new proof of the first inequality of (1.4. Main results Our method (which we discovered in fact in 010 [5] is based on the auiliary functions ( ( a 1 ( arccos, 0, π and ( a ( arccosh, > 0. In the proof, the well-known L Hospital rule for monotonicity lemma will be used, as follows (see e.g. []. Lemma.1. Suppose that f, g : (a, b R are differentiable on (a, b, and g ( 0 for (a, b. Suppose that f(a+ g(a+ 0. Then, if f is monotone on (a, b, then f is also g g monotone, having the same type of monotonicity. 15

3 Theorem.1. The function f 1 ( 1 ( arccos is strictly decreasing for ( ( Proof. We will use successively Lemma.1. Put f( arccos, g(. Then f ( g ( cos, and f(0+ g(0+ 0. sin ( f ( To avoid the radicals, we will consider u(0+ v(0+ 0 and u ( v ( and > cos. As u ( v ( 0, consider u ( v ( continuing, we obtain finally u ( v ( Now, remark that B( A( tan on ( ( cos ( sin u( v(. Here g ( sin cos sin cos clearly v ( 0 as > cos sin 1 + sin + 4 cos cos + sin + sin + 4 cos 4 cos 8 sin + 4 cos cos + cos A( B(. + 1, and it is well-known that tan cos + sin cos + sin. By is strictly increasing This means that A( is strictly decreasing, and the proof of Theorem.1. is finished. B( Theorem.. The function f ( 1 ( arccosh, > 0 is strictly increasing on (0,. ( Proof. Apply again Lemma.1 for f( arccosh and g(. As f(0+ g(0+ 0 and f ( g ( cosh, we will consider again sinh ( f ( g ( ( cosh (sinh u( v( and all can be repeated as in the proof of Theorem.1. One arrives finally to the function A( B( cosh B( and remark that in this case + cosh A( tanh + 1, which is strictly decreasing, as the function tanh is known to be strictly decreasing for > 0. Thus A( will be strictly increasing, so ( f ( g ( and ( f( g( B( will be strictly increasing, too. This last assertion implies also that f( is strictly increasing, too. g( ( Theorem.. The function f ( arccos is strictly decreasing on ( 16

4 Proof. Remark that f ( (f 1 ( 1, where f 1 ( is the decreasing function of Theorem.1. On the other hand, f 1 ( 1 h( is also decreasing, but h( < 0. This follows by > cos so the arccos function being decreasing, we get f 1 ( < 1. Now f ( h(, so f ( h( + h ( < 0 as h( < 0 and h ( < 0, > 0. ( Theorem.4. The function f 4 ( arccosh is strictly decreasing. Proof. The method of proof of Theorem. cannot be applied here, since f 4 ( (f ( 1, where f ( is the strictly increasing function of Theorem.. Now, f ( 1 < 0 as < cosh and the function arccosh is strictly increasing. By letting k( f ( 1 < 0, strictly increasing and f 4( k( + k (, where k( < 0, but k ( > 0. We will compute f 4( cosh sinh sinh. Now b( cosh sinh < 0 is equivalent to ( cosh < (sinh, or c( sinh cosh < 0. One has c ( 4( sinh < 0, after some elementary computations. Thus it follows c( < c(0+ 0, implying b( < 0, i.e. f 4( < 0. This finishes the proof of the theorem. Corollary.1. The best constants a, b > 0 such that for ( 0, π cos b <, are b 1, a π arccos π. < cos a, (.1 ( π ( π Proof. By Theorem.1 one has f 1 < f 1 ( < f 1 (0+. As f 1 π arccos π, and f 1 (0+ 1 (we will prove below, and the function cos being strictly decreasing, inequality ( π (.1 follows with best constants a f 1, b f 1 (0+. Now, to compute b lim f 1 (, 0 put y and as y 1 as 0+, remark that arccos y 1 as z arccos y sin(arccos y z 0 and sin z 1. Put sin(arccos y 1 y 1 sin. Therefore, as arccos y arccos y sin(arccos y, it is sufficient to compute the limit of 1 1 sin, or avoiding sin(arccos y sin the radicals, the limit of. 4 sin Now it is immediate by L Hospital s rule that 1 4, as sin ( ( +, and +, 17 1, so the limit follows. 6

5 Corollary.. The best constants c, d such that cos( + c < < cos( + d (. are for ( 0, π, c 0, d arccos π π ( Proof. Applying Theorem., we get f π ( < f ( < f (0+. Now, f π arccos ( π, π while f (0+ 0. Thus we get d < arccos < c, and applying the decreasing function cos, we get (.. Corollary.. The best constant k > 0 such that for > 0 is k 1. > cosh k, (. Proof. By Theorem. we get f ( > f (0+. Now, ( it follows by the same lines as in the 1 proof of Corollary.1 that f (0+ lim 0+ arccosh 1. As the function cosh is strictly increasing, relation (. follows. Remark.1. If (0, 0, the the best constant l > 0 such that > 0, will be < cosh l, (.4 ( 0 l arccosh 0 Corollary.4. The best constant s such that < cosh( + s (.5 for > 0, is s 0. ( Proof. By Theorem.4 one has f 4 ( < f 4 (0+ s, so arccosh < + s, implying < cosh( + s, with best s 0. Remark.. If (0, 0, the best constant p such that > cosh( + p (.6 for > 0, is p arccosh

6 References [1] Iyengar, K. S. K., Madhava Rao, B. S. & Nanjundiah, T. S. (1945 Some trigonometrical inequalities, Half-yearly J. Mysore Univ. B(N.S., 6, 1 1. [] Mitrinović, D. S. (1970 Analytic Inequalities, Springer Verlag, Berlin. [] Pinelis, I. (00 L Hospital type rules for monotonicity, with applications, J. Ineq. Pure Appl. Math., (1, article 5 (electronic. [4] Sándor, J. (011 Trigonometric and hyperbolic inequalities, arxiv: v1, 1 9. [5] Sándor, J. (010 Unpublished manuscripts. [6] Sándor, J. (017 Refinements of the Mitrinović Adamović inequality, and an application, Notes on Number Theory and Discrete Mathematics, (1, 4 6. [7] Sándor, J. (017. Two Applications of the Hadamard Integral Inequality, Notes on Number Theory and Discrete Mathematics, (4,