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[不等式] 關於三角形的不等式$\frac{1}{(\sqrt{r_a}+\sqrt{r_b})^2}$

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tommywong Posted 2019-7-12 18:53 |Read mode
Last edited by tommywong 2019-7-12 19:18證明 $\displaystyle \sum_{cyc}{\frac{1}{(\sqrt{r_a}+\sqrt{r_b})^2}}\leq\frac{1}{4r}$

(teomihai冇講啲符號喺乜,我諗應該係旁切圓半徑同內切圓半徑)
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kuing Posted 2019-7-12 21:25
完全就是在看你知不知道有
\[\frac1{r_a}+\frac1{r_b}+\frac1{r_c}=\frac1r\]而已,熟悉这方面知识的都能秒……

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