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..

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