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[不等式] $(1+\frac12+\dots+\frac1n)^2<n(1+\frac1{2^2}+\dots+\frac1{n^2})$

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APPSYZY posted 2023-10-5 19:07 |Read mode
Last edited by kuing 2023-11-8 16:18求证:
\[\left(1+\frac12+\dots+\frac1n\right)^2<n\left(1+\frac1{2^2}+\dots+\frac1{n^2}\right).\]

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爪机专用 posted 2023-10-5 19:08
Last edited by kuing 2023-11-8 16:16这不是柯西一下就好了嘛

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isee posted 2023-10-6 20:07
Last edited by isee 2023-11-7 11:58用均值不等式\[\sqrt{\frac{a_1^2+a_2^2+\cdots+a_n^2}n}\geqslant\frac{a_1+a_2+\cdots+a_n}n.\]

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😁😁😁  posted 2023-10-6 22:14

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isee=freeMaths@知乎

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