An Answer to the Conjecture of Satnoianu∗

An Answer to the Conjecture of Satnoianu∗

Applied Mathematics E-Notes, 9(2009), 262-265 c ISSN 1607-2510 Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/ An Answer To The Conjecture Of Satnoianu∗ Yu Miao†, Shou Fang Xu‡, Ying Xia Chen§ Received 23 September 2008 Abstract In this short paper, we obtain an answer to the conjecture of Satnoianu by a simpler method in the view of probability theory. The conditions of our results are independent with some known answers. 1 Introduction In [2], Mazur proposed the open problem: if a,b,c are positive real numbers such that abc > 29, then 1 1 1 3 + + . (1) √1 + a √1 + b √1 + c ≥ 1 + √3 abc In fact, in 2001, Satnoianu [3] has studied the followingp inequality a 3 (a,b,c> 0, λ 8). (2) √a2 + λbc ≥ √1 + λ ≥ cyclicX In addition, Satnoianu proposed the following inequality as a conjecture 1 n n−1 n−1 x − 1 i n(1 + λ) n−1 . (3) n−1 ≥ =1 xi + λ k6=i xk ! Xi Shortly after the proposed conjecture,Q Janous [1] gave the proof of the inequality (3) by means of Lagrange’s method of multipliers and Satnoianu [4] obtained a generalized version of inequality (3) as follows 1 n n−1 n−1 x − 1 i n(α + β) n−1 , (4) n−1 ≥ i=1 αxi + β k6=i xk ! X ∗Mathematics Subject Classifications:Q 26D15 †College of Mathematics and Information Science, Henan Normal University, Xinxiang, Henan, 453007, P. R. China. E-mail: [email protected] ‡Department of mathematics, Xinxiang University, Xinxiang, Henan, 453000, P. R. China §College of Mathematics and Information Science, Pingdingshan University, Pingdingshan, Henan, 467000, P. R. China 262 Miao et al. 263 n−1 where n 2, xi > 0, i = 1, 2,...,n, α,β > 0 and β (n 1)α. Recently, Wu [5] established≥ the following more generalized inequality≥ − 1 n q p x − 1 i n(α + β) p , (5) q n q/n ≥ i=1 αx + β x ! X i k=1 k where α,β,xi(i = 1, 2,...,n) are positiveQ real numbers, q R, and p < 0, or p > 0 with β (nmax{p,1} 1)α. ∈ If we≥ rewrite the− inequality (5) as 1 n p 1 1 − 1 (α + β) p , (6) n α + β exp 1 n log xq log xq ≥ i=1 n k=1 k i ! X − then it is easy to see that (6) is equivalentP to 1 p X − 1 E (α + β) p , (7) αX + β exp E log X ≥ { } q q q where X is a random variable taking values x1,x2,...,xn with the probability P (X = q 1 xi ) = n and E(X) denotes the mathematical expectation of X. In fact, X can be an any positive random variable. Hence we could generalize the conjecture of Satnoianu as: ”Under what conditions does the inequality (7) holds?” 2 Main Results Before our works, we need give the following useful LEMMA 1. Let f(x) = (a + bex)p, where a,b > 0, x R. If p > 0 or if p < 0 with pbex + a 0, then f(x) is a convex function. ∈ ≤ PROOF. The method is elementary. Since a twice differentiable function of one variable is convex on an interval if and only if its second derivative is non-negative and f 0(x) = pb(a + bex)p−1ex, f 00(x) = p(p 1)b2(a + bex)p−2e2x + pb(a + bex)p−1ex − = pbex(a + bex)p−2[(p 1)bex + (a + bex)] − = pbex(a + bex)p−2[pbex + a], the desired result is easy to be obtained. PROPOSITION 1. Let random variable X > 0 a.e. and α,β > 0. If p < 0 or if p > 0 with X βeE log X /(αp) a.e., then we have ≤ 1 p 1 X − E (α + β) p . (8) αX + β exp E log X ≥ { } 264 Conjecture of Satnoianu PROOF. Let Y = log X, then (8) is equivalent to − 1 p 1 − 1 E (α + β) p . (9) α + βe−EY eY ≥ By Lemma 1. and Jensen’s inequality, the proof is easy to be obtained. From the above proposition, we have the following result and the proof is easy. THEOREM 1. Let α,β > 0 and X be a discrete random variable taking positive n numbers x1,x2,...,xn with P (X = xi) = ai, where i=1 ai = 1. In addition, let M = max xi, 1 i n and m = min xi, 1 i n . If p < 0 or if p > 0 with M/m β/{(αp),≤ then≤ we} have { ≤ ≤ P} ≤ 1 n p xi − 1 a (α + β) p . (10) i αx + β n xai ≥ =1 i k=1 k Xi 1 In particular, if a1 = a2 = = a = Q, we have ··· n n 1 n p xi − 1 n(α + β) p . (11) n 1/n ≥ =1 αxi + β x ! Xi k=1 k REMARK 1. By comparing theQ conditions of Theorem 1. with the ones of Wu in [5], we find that these assumptions are independent each other. In fact, the only difference is between “M/m β/(αp)” and “β (nmax{p,1} 1)α”, from that we can not judge which condition is≤ weaker than the other.≥ − REMARK 2. For the infinite sequence x ∞ , let ∞ a = 1, M = sup x < { i}i=1 i=1 i i≥1 i and m = infi≥1 xi > 0, then by the same discussions as Theorem 1., we have ∞ P 1 ∞ p xi − 1 a (α + β) p . (12) i αx + β ∞ xai ≥ =1 i k=1 k Xi The following result is the integralQ form of the conjecture of Satnoianu. THEOREM 2. Let α,β > 0 and X be a positive continuous random variable on (0, ) with the probability density function f(x). If p < 0 or if p > 0 with X βe∞E log X /(αp) a.e., then we have ≤ 1 ∞ p x 1 − p ∞ f(x)dx (α + β) . (13) αx + β exp log xf(x)dx ≥ Z0 0 ! In particular, if X possesses uniformR distribution on the support interval [a,b], i.e., the probability density function of X is equal to (b a)−1,x [a,b] and zero elsewhere. Then if b/a β/(αp), then we have − ∈ ≤ 1 p b 1 x − 1 dx (α + β) p . (14) b b a a 1 ≥ − Z αx + β exp b−a a log xdx n R o Miao et al. 265 References [1] W. Janous, On a conjecture of Razvan Satnoianu, Gazeta Matematica Seria A, XX(XCIX) (2002), 97–103. [2] M. Mazur, Problem 10944, Amer. Math. Monthly, 109(2002), 475. [3] R. A. Satnoianu, The proof of the conjectured inequality from theMathematical Olympiad, Washington DC 2001, Gazeta Matematica, 106 (2001), 390–393. [4] R. A. Satnoianu, Improved GA-convexity inequalities, J. Inequal. Pure Appl. Math., 3(5)(2002), Art. 82. [5] S. H. Wu, Note on a conjecture of R. A. Satnoianu, Math. Inequal. Appl. 12(1)(2009), 147–151..

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    4 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us