$ \left\{\begin{array}{l} \sum\limits_{i=1}^{1999}x_i=1999\\\sum\limits_{i=1}^{1999}x_i^3 = \sum\limits_{i=1}^{1999}x_i^4\end{array}\right.$
Edited by pirate, 14-07-2007 - 08:02.
#1
Posted 14-07-2007 - 08:02
pirate
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Trung sĩ
Giải hệ phương trình:
$ \left\{\begin{array}{l} \sum\limits_{i=1}^{1999}x_i=1999\\\sum\limits_{i=1}^{1999}x_i^3 = \sum\limits_{i=1}^{1999}x_i^4\end{array}\right.$
$ \left\{\begin{array}{l} \sum\limits_{i=1}^{1999}x_i=1999\\\sum\limits_{i=1}^{1999}x_i^3 = \sum\limits_{i=1}^{1999}x_i^4\end{array}\right.$
#2
Posted 15-07-2007 - 18:57
hoang tuan anh
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^^
áp dụng chebyshev cho 2 dãy đơn điệu cùng chiều $(x_i); (x_i^3)$
ta có $\sum^{1999}_{i=1}x_i^4 \geq \dfrac{1}{1999}(\sum^{1999}_{i=1} x_i)(\sum^{1999}_{i=1} x_i^3)$
ta có $\sum^{1999}_{i=1}x_i^4 \geq \dfrac{1}{1999}(\sum^{1999}_{i=1} x_i)(\sum^{1999}_{i=1} x_i^3)$
HTA
dont put off until tomorrow what you can do today
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