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很经典的题目, 论坛里肯定出现过.
注意到对任意 $\lambda,\mu>0$, 总有
\begin{align*}
\left(\sum_{n\geq 1} a_n\right)^2&=\left(\sum_{n\geq 1} (\lambda+\mu n^2)^{-1}\right)\left(\sum_{n\geq 1} (\lambda+\mu n^2)a_n^2\right)\\
&\leq \int_0^\infty \dfrac{\mathrm dt}{\lambda +\mu t^2}\cdot \left(\lambda\sum_{n\geq 1}a_n^2+\mu \sum_{n\geq 1} n^2a_n^2\right)\\
&=\dfrac{\pi}{2\sqrt{\lambda \mu}}\cdot \left(\lambda\sum_{n\geq 1}a_n^2+\mu \sum_{n\geq 1} n^2a_n^2\right).
\end{align*}
取 $\lambda^{-1}=\mu =\sqrt{\dfrac{\sum_{n\geq 1}a_n^2}{\sum_{n\geq 1}n^2a_n^2}}$ 即可. |
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Czhang271828
posted 2023-4-26 19:30
