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[不等式] 一个不精细的不等式

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力工 posted 2023-7-24 22:12 |Read mode
这题还是很不细的,据传是中科大招生题:
证明:$\dfrac{1^2}{n^2+1^2}+\dfrac{2^2}{n^2+2^2}+\cdots \dfrac{n^2}{n^2+n^2}\leqslant \frac{n^2+2n}{4n+2}$

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O-17 posted 2023-7-25 01:37
\begin{align*}
\sum_{k=1}^{n}\frac{k^2}{n^2+k^2}\leqslant\frac{n^2+2n}{4n+2}\Leftrightarrow{}&\sum_{k=1}^{n}\left(1-\frac{k^2}{n^2+k^2}\right)\geqslant n-\frac{n^2+2n}{4n+2}\\\Leftrightarrow{}&\sum_{k=1}^{n}\frac{1}{n^2+k^2}\geqslant\frac3{4n+2}\\\Leftarrow{}&\frac{n^2}{\sum_{k=1}^{n}\left(n^2+k^2\right)}\geqslant\frac{3}{4n+2}~~\text{(AM-HM)}\\\Leftrightarrow{}&\frac{n^2}{n^3+\frac{n(n+1)(2n+1)}{6}}\geqslant\frac{3}{4n+2}\\\Leftrightarrow{}&\frac{2}{8n+\frac1n+3}\geqslant\frac1{4n+2}\\\Leftrightarrow{}&n\geqslant1
\end{align*}

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