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Last edited by hbghlyj 2021-2-20 15:10来自qianxiangzhen3
$\triangle ABC$的内角均小于$\frac{2\pi}3$
r是内切圆半径,费马点R的赛瓦三角形DEF的外接圆的半径$<\frac2{\sqrt3}r$
AR=x,BR=y,CR=z
等价于
$\frac{\sum y z}{\sum\sqrt{y^2+y z+z^2}}\ge\frac{\sqrt{\prod\left(y^2 (x+z)^2+z^2 (x+y)^2+y z (x+y) (x+z)\right)}}{2\sqrt3\cdot \sum yz\prod (y+z)}$
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