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设△ABC的内角均小于120°。在三角形内可以找到一点P,使∠APB=∠BPC=∠CPA=120°,该点到三顶点的距离之和小于其它任何点到三顶点的距离之和,该点称为费尔玛极值点。下面借助费尔玛极值点证明一个不等式: a~2+b~2+c~2≥4 3~(1/2)△。(a、b、c为三边长,△为三角形的面积)。设PA,PB,PC依次为r_1,r_2,r_3。则 a~2+b~2+c~2=2r_1~2+2r_2~2+2r_3~2-2r_1r_2cos120°-2r_2r_3cos120°-2r_3r_1cos120°=2(r_1~2+r_2~2+r_3~2)
Let ΔABC have interior angles less than 120°. A point P can be found in a triangle such that ∠APB=∠BPC=∠CPA=120°. The sum of the distances from this point to the three vertices is smaller than the sum of the distances from any other point to the three vertices. This point is called the Fairchild pole. Value point. An inequality is proved by using the Fairmax extremum: a~2+b~2+c~2≥4 3~(1/2)△. (a, b, c are three sides long, △ is the area of the triangle). Let PA, PB, PC be r_1, r_2, r_3 in order. Then a~2+b~2+c~2=2r_1~2+2r_2~2+2r_3~2-2r_1r_2cos120°-2r_2r_3cos120°-2r_3r_1cos120°=2(r_1~2+r_2~2+r_3~2)