a^2+b^2+c^2=1 用高等数学中求多元函数极值的方法求ab+ac+bc的最大最小值。如何解驻点，驻点如何取舍？

先列出方程组: b+c+2aλ = 0, a+c+2bλ = 0, a+b+2cλ = 0, a²+b²+c² = 1. 前三式两两相减得: (2λ-1)(a-b) = 0, (2λ-1)(b-c) = 0. 若2λ-1 ≠ 0, 则a-b = 0, b-c = 0, 即a = b = c. 代回解得a = b = c = ±1/√3, λ = -1, 对应ab+bc+ca = 1. 可验证函数ab+bc+ca-(a²+b²+c