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[几何] 四面体内任意点的几何不等式

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lihpb Posted 2023-11-4 22:19 |Read mode
Last edited by hbghlyj 2023-11-5 15:21$P$为四面体$A_1A_2A_3A_4$内任意点。
$A_iP$连线交$A_i$所对侧面$S_i$于$B_i$。
令$A_iP=X_i$,$B_iP=Y_i$,则
\[X_1X_2X_3X_4≥81Y_1Y_2Y_3Y_4.\]
等号成立的充要条件是$P$为四面体重心。

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kuing Posted 2023-11-4 23:29
Last edited by kuing 2023-11-5 21:48咦,这个简单😁

记四面体体积为 `V`,由 `P` 与侧面 `S_i` 连成的体积为 `V_i`,再令 `X_i=t_iY_i`,则
\[\frac{V_i}V=\frac{Y_i}{X_i+Y_i}=\frac1{t_i+1},\]
所以
\[\frac1{t_1+1}+\frac1{t_2+1}+\frac1{t_3+1}+\frac1{t_4+1}=1,\]
待证的就是 `t_1t_2t_3t_4\geqslant81`,再令
\[\frac1{t_1+1}=\frac a{a+b+c+d}\]
等,得
\[t_1=\frac{b+c+d}a\]
等,待证的就变成
\[(b+c+d)(c+d+a)(d+a+b)(a+b+c)\geqslant81abcd,\]
显然成立。

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 Author| lihpb Posted 2024-6-6 17:37
kuing 发表于 2023-11-4 23:29
咦,这个简单😁

记四面体体积为 `V`,由 `P` 与侧面 `S_i` 连成的体积为 `V_i`,再令 `X_i=t_iY_i`,则
帮忙证下这个,有点像
forum.php?mod=viewthread&tid=12440

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