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

APPSYZY · 2023-10-5 19:07

Last edited by kuing 2023-11-8 16:18求证：
\[\left(1+\frac12+\dots+\frac1n\right)^2<n\left(1+\frac1{2^2}+\dots+\frac1{n^2}\right).\]

爪机专用 · 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

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