悠闲数学娱乐论坛(第3版)»论坛 › 数学区 › 初等数学讨论 › $\frac1a<\frac{2π\tan \left(\frac12π\sqrt{1-8 a}\right)}{\sqrt{1-8 a}}$

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[不等式] $\frac1a<\frac{2π\tan \left(\frac12π\sqrt{1-8 a}\right)}{\sqrt{1-8 a}}$

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hbghlyj Posted at 2025-3-1 00:42:49 |Read mode

$0<a<\frac18:$\[\frac1a<\frac{2 \pi \tan \left(\frac{1}{2} \pi \sqrt{1-8 a}\right)}{\sqrt{1-8 a}}\]

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神奇的是这个8还是最大的 Posted at 2025-3-1 21:44

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Author| hbghlyj Posted at 2025-3-1 03:25:22

刚刚回帖遇到“不存在”的问题，非常令人困惑

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🥺 Posted at 2025-3-1 13:37

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