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[几何] 一道有关塞瓦三角形的几何不等式

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1+1=? posted 2025-7-13 01:48 |Read mode
Last edited by hbghlyj 2025-7-13 04:12如下两个结论:$P$为$\triangle ABC$内一点,$P$点的Ceva三角形为$\triangle DEF$,记$\triangle ABC$,$\triangle DEF$的面积分别为$S_1$,$S_2$,则有如下不等式:$$\dfrac{PD}{PA}+\dfrac{PE}{PB}+\dfrac{PF}{PB}\geqslant  3(\frac{S_2}{2S_1})^{\frac{1}{3}}$$

对任意$\triangle ABC$,设$R$,$r$分别为$\triangle ABC$的外接圆与内接圆半径,则有:$$2-\dfrac{2r}{R} \leqslant \sum\tan^2 \dfrac{A}{2} \leqslant \dfrac{\dfrac{9}{2}R^2}{\sum bc}$$

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