Abstract
Let (GA) [k] n (a), A n (a), G n (a) be the third symmetric mean of k degree, the arithmetic and geometric means of a 1, …, a n (a i > 0, i = 1, …, n), respectively. By means of descending dimension method, we prove that the maximum of p is k−1/n−1 and the minimum of q is n/n−1(k−1/k)k/n so that the inequalities {fx505-1} hold.
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Wen, J., Yuan, J. & Yuan, S. An optimal version of an inequality involving the third symmetric means. Proc Math Sci 118, 505–516 (2008). https://doi.org/10.1007/s12044-008-0038-0
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DOI: https://doi.org/10.1007/s12044-008-0038-0