Question

已知实数a,b,c满足a^2+b^2=4 b^2+c^2=8 则ab+bc+2／根号2ac的最小值为

求详细过程

Answer 1

由a, b, c为实数, a²+b² = 4, b²+c² = 8,
有0 ≤ (a+√2·b+c)² = a²+2b²+c²+2√2·ab+2√2·bc+2ca = 12+2√2(ab+bc+√2/2·ca).
故ab+bc+√2/2·ca ≥ -12/(2√2) = -3√2.
等号成立当且仅当a+√2·b+c = 0.

下面说明方程组a²+b² = 4, b²+c² = 8, a+√2·b+c = 0在实数范围内有解.
只需说明√(4-b²)+√2·b+√(8-b²) = 0在[-2,2]中有解.
由f(x) = √(4-x²)+√2·x+√(8-x²)连续, 又f(-2) = 2-2√2 < 0, f(0) = 2+2√2 > 0,
f(x) = 0在(-2,0)中有解.

于是ab+bc+√2/2·ca ≥ -3√2能够成立等号, ab+bc+√2/2·ca的最小值就是-3√2.