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第一届全国几何不等式研讨会上,刘健和刘毅老师曾提出如下的 猜想 设△ABC的三条内角平分线的长分别为t_a,t_b,t_c,其外接圆与内切圆的半径分别是R与r,则 t_a~2+t_b~2+t_c~2≥(27/2)Rr. (1) 当且仅当△ABC为正三角形时等号成立. 本文将对(1)给出肯定的回答。 证明 设BC=a,CA=b,AB=c,且△ABC的面积为S,则有
At the 1st National Symposium on Geometric Inequalities, Liu Jian and Liu Yi proposed the following hypothesis: The lengths of the three internal angle bisectors of △ABC are t_a, t_b, and t_c, respectively. The radii of the circumcircle and inscribed circle are respectively R and r, then t_a~2+t_b~2+t_c~2≥(27/2)Rr. (1) If and only if △ABC is an equilateral triangle, the equal sign is established. This article will give a positive answer to (1). The answer. Proof Let BC=a, CA=b, AB=c, and the area of △ABC is S, then there is