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熟知,a·b≤|a||b|,当且仅当向量a与b方向相同时等号成立.利用这一性质,可以轻松解决一类变化多样的问题.例1设x、y∈R~+,且x+y=1,求证:(x+1)~(1/2)+(y+2)~(1/2)≤2(2)~(1/2)解解法一(不等式法):
It is well-known that a · b ≤ | a || b |, and if and only if the vectors a and b are in the same direction, the equals sign holds. Using this property, we can easily solve a wide variety of problems. (X + 1) ~ (1/2) + (y + 2) ~ (1/2) ≤2 (2) ~ (1/2) (Inequality method):