Cho các số thực $a_1,a_2,...,a_n \in [p,q] (p,q \ge 0, n$ là số chẵn $)$. Chứng minh
$(a_1+a_2+...+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}) \leq n^2 + \frac{n^2.(p-q)^2}{4pq}$
Edited by quangtq1998, 20-07-2013 - 15:38.
$(a_1+a_2+...+a_n)(\frac{1}{a_1}++\frac{1}{a_2}+...+\frac{1}{a_n}) \leq n^2 + \frac{n^2.(p-q)^2}{4pq}$
Started By quangtq1998, 20-07-2013 - 15:36
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Posted 20-07-2013 - 15:36
quangtq1998
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