On Toader-Sándor Mean 1

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1 International Mathematical Forum, Vol. 8, 213, no. 22, HIKARI Ltd, On Toader-Sándor Mean 1 Yingqing Song College of Mathematics and Computation Sciences Hunan City University Yiyang, Hunan, 413, P.R. China Boyong Long School of Mathematics Sciences Anhui University Hefei, Anhui, 2339, P. R. China Yuming Chu College of Mathematics and Computation Sciences Hunan City University Yiyang, Hunan, 413, P.R. China chuyuming@hutc.zj.cn Copyright c 213 Yingqing Song et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper, we establish some inequalities for the Toader-Sándor mean by use of Jensen inequality and convexity. Mathematics Subject Classification: 26E6, 26D15 Keywords: Toader-Sándor mean, Jensen inequality, convexity 1. Introduction 1 This research was supported by the Natural Science Foundation of China under Grant and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T2924.

2 158 Yingqing Song, Boyong Long and Yuming Chu Recently, the means M(a, b) of two positive numbers a and b have been the subject of intensive research. In particular, many remarkable inequalities between different means can be found in the literature [1-26]. Let R be the real numbers set and R + the positive real numbers set, and J R + be an interval. If f : J R be a strictly monotone function and p : J R + be a positive function, then the weighted quasi-arithmetic integral mean M(f,p) [27] on J is defined by ( b M(f,p) =M(f,p)(a, b) =f 1 f(x)p(x)dx ) a b p(x)dx, a, b J. (1.1) a The weighted quasi-arithmetic integral mean M(f,p) was considered in [28] for arbitrary weight function p and special f defined by f(x) = { x n, n, log x, n =. More means of type M(f,p) are given in [29, 3], but only for special cases of functions f or p. In [31], Sándor and Toader proved that inequality M(f,p) <M(g, p) holds for any p if f : R + R + is strictly monotone, g : R + R + is strictly increasing, and the composed function g f 1 is convex. Making use of two functions but only one integral, Toader and sándor [32] defined another integral mean N(f,p) as follows. Let f : J R and p : J R be two strictly monotone functions on J. Then the Toader-Sándor mean N(f,p) [8] on J is defined by ( ) N(f,p) =N(f,p)(a, b) =f 1 (f p 1 )[t p(a)+(1 t) p(b)]dt. (1.2) It is not difficult to verify that if the function p is differentiable with p >, then N(f,p) =M(f,p ). (1.3) In [33], the Schur convexity and concavity for M(f,p ) were discussed. The following Theorems 1.1 and 1.2 can be found in [32]. Theorem 1.1. If the function f : R + R + is strictly monotone, the function g : R + R + is strictly increasing, and the composed function g f 1 is convex, then the inequality N(f,p) <N(g, p) holds for every monotone function p.

3 On Toader-Sándor mean 159 Theorem 1.2. Ifp is a strictly monotone real function on J and q is a strictly increasing real function on J, such that q p 1 is strictly convex, then N(f,p) <N(f,q) for each strictly monotone function f. The following Lemmas 1.3 and 1.4 were established by Hutník in [34]. Lemma 1.3. Let f :[a, b] [c, d] and let f 1 be the inverse function to the function f. (1) If f is strictly increasing and convex, or a strictly decreasing and concave function on [a, b], then f 1 is a concave function on [c, d]. (2) If f is strictly decreasing and convex, or a strictly increasing and concave function on [a, b], then f 1 is a convex function on [c, d]. Lemma 1.4. Let ϕ :[a, b] [c, d] and h :[c, d] R. (1) If ϕ is convex on [a, b] and h is convex increasing on [c, d], or ϕ is concave on [a, b] and h is convex decreasing on [c, d], then h ϕ is convex on [c, d]. (2) If ϕ is convex on [a, b] and h is concave decreasing on [c, d], or ϕ is concave on [a, b] and h is concave increasing on [c, d], then h ϕ is concave on [c, d]. In general measure theoretical notation the Jensen inequality theorem sounds as follows: let (Ω,A,μ) be a measurable space, such that μ(ω) = 1. If f is a real μ-integrable function and φ is a convex (concave) function on the range of f, then ( ) φ fdμ ( ) φ fdμ. Ω Ω The purpose of this paper is to establish some new and interesting inequalities for the Toader-Sándor mean N(f,p). 2. Main Result Theorem 2.1. Let f and p be two strictly monotone functions, and A(a, b) = (a + b)/2 the arithmetic mean. Then the following statements are true. (1) If f 1 and p 1 are convex, then N(f,p)(a, b) A(a, b). (2.1) (2) If f 1 and p 1 are concave, then inequality (2.1) is reversed. Proof. We only give the proof of inequality (2.1) in detail, proof of the remaining part is similar. If f 1 is convex, then Jensen inequality implies that N(f,p) = f 1 ( (f p 1 )[t p(a)+(1 t) p(b)]dt p 1 [t p(a)+(1 t) p(b)]dt = N(I,p), (2.2) )

4 16 Yingqing Song, Boyong Long and Yuming Chu where I = I(t) = t denotes the identity transformation. If p 1 is convex, then N(I,p)= p 1 [t p(a)+(1 t) p(b)]dt [ta +(1 t)b]dt = a + b 2 = A(a, b). (2.3) Therefore, inequality (2.1) follows from inequalities (2.2) and (2.3). Theorem 2.2. Let p be a strictly monotone function, and f a strictly increasing function. (1) If f is concave, then N(f,p) N(f 1,p). (2.4) (2) If f is convex, then inequality (2.4) is reversed. Proof. Iff is strictly increasing and concave, then by Lemma 1.3(2) we know that f 1 is convex. Therefore, inequality (2.2) holds again and it follows from Jensen inequality that f 1 1 p 1 [t p(a)+(1 t) p(b)]dt (f 1 p 1 )[t p(a)+(1 t) p(b)]dt. Moreover, if f is strictly increasing, then the above inequality leads to N(I,p)= p 1 [t p(a)+(1 t) p(b)]dt 1 f (f 1 p 1 )[t p(a)+(1 t) p(b)]dt = N(f 1,p). (2.5) Therefore, inequality (2.4) follows from inequalities (2.2) and (2.5). The proof of part (2) is similar. Theorem 2.3. Let f and p be two strictly monotone functions and g defined on the range of f. Then the following statements are true. (1) If g is convex and g f is strictly increasing, or g is concave and g f is strictly decreasing, then N(f,p) N(g f,p). (2.6) (2) If g is concave and g f is strictly increasing, or g is convex and g f is strictly decreasing, then inequality (2.6) is reversed.

5 On Toader-Sándor mean 161 Proof. Ifg is convex, then the Jensen inequality implies that 1 g (f p 1 )[t p(a)+(1 t) p(b)]dt (g f p 1 )[t p(a)+(1 t) p(b)]dt. If g f is strictly increasing, then (g f) 1 exists and is strictly increasing. By composition with (g f) 1 on the both sides of the above inequality, one has f 1 1 (f p 1 )[t p(a)+(1 t) p(b)]dt (g f) 1 ( (g f p 1 )[t p(a)+(1 t) p(b)]dt i.e. inequality (2.6) holds. The proofs of remaining parts are similar. Theorem 2.4. Let f and p be two strictly monotone functions. Then we have (1) If f g f 1 is convex and f g is strictly increasing, or f g f 1 is concave and f g is strictly decreasing, then ), N(f,p) N(f g, p). (2.7) (2) If f g f 1 is convex and f g is strictly decreasing, or f g f 1 is concave and f g is strictly increasing, then inequality (2.7) is reversed. Proof. Iff g f 1 is convex, then the Jensen inequality implies that (f g f 1 1 ) (f p 1 )[t p(a)+(1 t) p(b)]dt (f g p 1 )[t p(a)+(1 t) p(b)]dt. If f g is strictly increasing, then (f g) 1 exists and is strictly increasing. By composition with (f g) 1 on the both sides of the above inequality, we have f 1 1 (f p 1 )[t p(a)+(1 t) p(b)]dt (f g) 1 ( (f g p 1 )[t p(a)+(1 t) p(b)]dt i.e., inequality (2.7) holds. The proofs of the remaining parts are similar. Theorem 2.5. Let f and p be two strictly monotone functions, then the following statements are true: ),

6 162 Yingqing Song, Boyong Long and Yuming Chu (1) If function g is strictly decreasing and g f 1 is concave, then N(f,p) N(g, p). (2.8) (2) If function g is strictly decreasing and g f 1 is convex, or g is strictly increasing and g f 1 is concave, then inequality (2.8) is reversed. Proof. The proof is similar to the proof of Theorem 1.1. For the reader s convenience, we only give the proof of inequality (2.8) in detail as follows. If g f 1 is concave, then by the Jensen inequality, one has (g f 1 1 ) (f p 1 )[t p(a)+(1 t) p(b)]dt (g p 1 )[t p(a)+(1 t) p(b)]dt. If g is strictly decreasing, then g 1 exists and is strictly decreasing. By composition with g 1 on the both sides of the above inequality, we get f 1 1 (f p 1 )[t p(a)+(1 t) p(b)]dt g 1 1 (g p 1 )[t p(a)+(1 t) p(b)]dt Therefore, inequality (2.8) holds. Theorem 2.6. Let f and p be two strictly monotone functions. (1) If function q is strictly decreasing and q p 1 is concave, then N(f,p) N(f,q). (2.9) (2) If function q is strictly increasing and q p 1 is concave, or q is strictly decreasing and q p 1 is convex, then inequality (2.9) is reversed. Proof. The proof is similar to that of Theorem 1.2. For convenience of the reader, we only give the proof of (2.9) in detail. Let p(a) =c, p(b) =d. Ifq p 1 is concave, then (q p 1 )[tc +(1 t)d] t(q p 1 )(c)+(1 t)(q p 1 )(d). If q is strictly decreasing, then q 1 exists and is strictly decreasing. By composition with q 1 on both sides of the above inequality, one has p 1 [tp(a)+(1 t)p(b)] q 1 [tq(a)+(1 t)q(b)]. (2.1) We divide the discuss into two cases. Case 1 f is strictly increasing. Then it follows from inequality (2.1) that (f p 1 )[tp(a)+(1 t)p(b)] (f q 1 )[tq(a)+(1 t)q(b)]. (2.11).

7 On Toader-Sándor mean 163 Inequality (2.11) leads to (f p 1 )[tp(a)+(1 t)p(b)]dt (f q 1 )[tq(a)+(1 t)q(b)]dt. (2.12) Composing with f 1 on both sides of inequality (2.12), we obtain the desired inequality (2.9). Case 2 f is strictly decreasing. Both inequalities (2.11) and (2.12) are reversed. But the inequality (2.9) holds again. Theorem 2.7. Let f be a strictly monotone function and p a strictly increasing function, and p :[a, b] [c, d] with [c, d] [a, b]. (1) If p is convex, then N(f,p 1 )(c, d) N(f,p)(c, d). (2.13) (2) If p is concave, then inequality (2.13) is reversed. Proof. We only give the proof of inequality (2.13) in detail. If p is convex, then p(t p 1 (c)+(1 t)p 1 (d)) tc +(1 t)d. (2.14) If p is strictly increasing and convex, then Lemma 1.3(1) implies that p 1 exists and is concave. Therefore, tc +(1 t)d = t(p 1 p)(c)+(1 t)(p 1 p)(d) p 1 (t p(c)+(1 t) p(d)). (2.15) Inequalities (2.14) and (2.15) implies that p(t p 1 (c)+(1 t)p 1 (d)) p 1 (t p(c)+(1 t) p(d)) (2.16) if p is strictly increasing and convex. We divide the discuss into two cases. Case 1 f is strictly increasing. Then it follows from (2.16) that (f p)(t p 1 (c)+(1 t)p 1 (d)) (f p 1 )(t p(c)+(1 t) p(d)). (2.17) It follows from (2.17) that (f p)(t p 1 (c)+(1 t)p 1 (d))dt (f p 1 )(t p(c)+(1 t) p(d))dt. (2.18)

8 164 Yingqing Song, Boyong Long and Yuming Chu Therefore, inequality (2.13) follows from (2.18). Case 2 f is decreasing. Then inequalities (2.17) and (2.18) are reversed. But inequality (2.13) holds again. Theorem 2.8. Let f and p be two strictly monotone functions and q a strictly decreasing function. Then the following statements are true. (1) If q is concave and p is convex and strictly decreasing, or q is concave and p is concave and strictly increasing, then N(f,p) N(f,p q). (2.19) (2) If q is convex and p is concave and strictly decreasing, or q is convex and p is convex and strictly increasing, then inequality (2.19) is reversed. Proof. If p is convex and strictly decreasing, or concave and strictly increasing, then Lemma 1.3(2) implies that p 1 is convex. Therefore, p 1 [tp(a)+(1 t)p(b)] ta +(1 t)b. (2.2) If q is concave and strictly decreasing, then it follows from Lemma 1.3(1) that q 1 is concave and decreasing. Because p 1 is convex, Lemma 1.4(2) leads to that q 1 p 1 =(p q) 1 is concave. Therefore, ta +(1 t)b (p q) 1 [t(p q)(a)+(1 t)(p q)(b)]. (2.21) From inequalities (2.2) and (2.21) we get p 1 [tp(a)+(1 t)p(b)] (p q) 1 [t(p q)(a)+(1 t)(p q)(b)]. (2.22) Therefore, inequality (2.19) follows easily from inequality (2.22). The proof of part (2) is similar. Theorem 2.9. Let f, p and q be three strictly monotone functions. (1) If p 1 is strictly increasing and q 1 is concave, or p 1 is strictly decreasing and q 1 is convex, then N(f,p) N(f,q p). (2.23) (2) If p 1 is strictly decreasing and q 1 is concave, or p 1 is strictly increasing and q 1 is convex, then inequality (2.23) is reversed. Proof. Ifq 1 is concave, then tp(a)+(1 t)p(b) q 1 [t(q p)(a)+(1 t)(q p)(b)]. It follows from the strict monotonicity of p 1 that p 1 [tp(a)+(1 t)p(b)] (q p) 1 [t(q p)(a)+(1 t)(q p)(b)]. (2.24) Therefore, inequality (2.23) follows easily from (2.24). The proofs of the remaining parts are similar.

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