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[不等式] 柯西不等式的简单应用

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Shiki posted 2020-9-9 22:46 |Read mode
Last edited by Shiki 2020-9-10 07:46对正数$a_i,b_i (i=1,2,\cdots,n)$
求证:
$$\sum^n_{i=1} \frac {a_i}{\sum^n_{k=1}a_kb_k - a_ib_i} \geqslant \frac {4}{\sum^n_{i=1}b_i}$$
= =

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kuing posted 2020-9-10 02:54
标题太短……连正数也不写……啧……罢了,反正题也是没啥意思……

记 `S=\sum_{k=1}^na_kb_k`,不等式就是
\[\sum_{i=1}^nb_i\sum_{i=1}^n\frac{a_i}{S-a_ib_i}\geqslant4,\]由 CS 及 AG 有
\[LHS\geqslant\left( \sum_{i=1}^n\sqrt{\frac{a_ib_i}{S-a_ib_i}} \right)^2=\left( \sum_{i=1}^n\frac{2a_ib_i}{2\sqrt{a_ib_i(S-a_ib_i)}} \right)^2\geqslant\left( \sum_{i=1}^n\frac{2a_ib_i}S \right)^2=4.\]

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original poster Shiki posted 2020-9-10 07:43
回复 2# kuing
让我完善一番

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