Báo cáo " The parameter-dependent cyclic inequality "

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In this paper we will construct a parameter-dependent cyclic inequality that can be used to prove a lot of hard and interesting inequalities. 1. Introduction The cyclic inequality is a type of inequality that may be right in just some particular cases but not in genenal. In this paper, we propose one type of parameter-dependent cyclic inequality from a special inequality. Thanks to this inequality, we can obtain many inequalities by choosing α and n. Note that it can be proved by some ways in particular case. However in order to prove it in general case, we have to...

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VNU Journal of Science, Mathematics - Physics 23 (2007) 155-158 The parameter-dependent cyclic inequality Nguyen Vu Luong∗ Department of Mathematics, Mechanics, Informatics, College of Science, VNU 334 Nguyen Trai, Hanoi, Vietnam Received 15 November 2006; received in revised form 12 September 2007 Abstract. In this paper we will construct a parameter-dependent cyclic inequality that can be used to prove a lot of hard and interesting inequalities. 1. Introduction The cyclic inequality is a type of inequality that may be right in just some particular cases but not in genenal. In this paper, we propose one type of parameter-dependent cyclic inequality from a special inequality. Thanks to this inequality, we can obtain many inequalities by choosing α and n. Note that it can be proved by some ways in particular case. However in order to prove it in general case, we have to use the method that is mentioned in the paper. 2. The general case Denote R+ = {x ∈ Rx > 0}. Lemma 1.1. Assume that xi ∈ R, (i = 1, n) we have n n−1 2 xi xj ≤ xi . 2n 1≤i
Nguyen Vu Luong / VNU Journal of Science, Mathematics - Physics 23 (2007) 155-158 156 Adding both sides of the above inequality by 2(n − 1) 1≤i
Nguyen Vu Luong / VNU Journal of Science, Mathematics - Physics 23 (2007) 155-158 157 Next, for n = 2k, we get x2 x2 1 2 P= + + 1 1 x2 x2 + α(x1 x2 + · · · + x1xk + x1 xk+1 ) + α(x2x3 + · · · + x2 xk+1 + x2 xk+2 ) 1 2 2 2 x2 n +···+ . 1 x2 + α(xn x1 + · · · + xn xk−1 + xn xk ) n 2 Applying the inequality (1.2), we get ( n=1 xi )2 n x i )2 ( i i=1 P = x2 + α 1≤i
Nguyen Vu Luong / VNU Journal of Science, Mathematics - Physics 23 (2007) 155-158 158 Take a = b, b = c we get Example 1.5. Assume that a, b ∈ R+ , α 2, prove that a((α + 2)a + 2αb) b((α + 2)b + 2αa) 2 + . [2a + α(2a + b)][2a + 3αb] [2b + α(2b + a)][2b + 3αa] 2 + 3α For n = 5 we yield the inequality Example 1.7. Give a, b, c, d, e ∈ R+ , α 2, prove that a b c d e 5 P= + + + + . a + α(b + c) b + α(c + d) c + α(d + e) d + α(e + a) e + α(a + b) 1 + 2α Take c = d = e, α = 2 we yield the inequality Example 1.8. Given a, b, c ∈ R+ , prove that a b 2a + 2c + b 4 + + 2c . a + 2b + 2c b + 4c [c + 2(c + a)][c + 2(a + b)] 5 For n = 6 we yield Example 1.9. Given ai ∈ R+ (i = 1, 6), α 2, prove that a1 a2 a3 + + + 1 1 1 a1 + α(a2 + a3 + a4) a2 + α(a3 + a4 + a5 ) a3 + α(a4 + a5 + a6) 2 2 2 a4 a5 a6 12 + + + 1 1 1 2 + 5α a4 + α(a5 + a6 + a1) a5 + α(a6 + a1 + a2 ) a6 + α(a1 + a2 + a3) 2 2 2 Finally, take a1 = a2 = a, a3 = a4 = b, a5 = a6 = c and α = 2 we get Example 1.10. Assume that a, b, c ∈ R+ , prove that 1 1 1 1 1 1 a + +b + +c + 1. 3a + 3b a + 4b + c 3b + 3c b + 4c + a 3c + 3a c + 4a + b Acknowledgements. This paper is based on the talk given at the Conference on Mathematics, Me- chanics, and Informatics, Hanoi, 7/10/2006, on the occasion of 50th Anniversary of Department of Mathematics, Mechanics and Informatics, Vietnam National University, Hanoi. References [1] B.A Troesch, The cyclic inequality for a large number of terms, Notices Amer. Math Soc. 25 (1978). [2] E.S Freidkin, S.A Freidkin, On a problem by Shapiro, Elem. Math. 45 (1990) 137.