论文部分内容阅读
结论设实数a、b满足a≥b,则对任意实数x有|x-a|+|x-b|≥a-b,当且仅当b≤x≤a时等号成立.证法1(零点分段讨论法)因为b,a分别是|x-b|和|x-a|的零点,于是分三种情况讨论:(1)当xa-b;(2)当b≤x≤a时,|x-a|+|x-b|=(a-x)+(x-b)=a-b;(3)当x>a时,|x-a|+|x-b|=(x-a)+(x-b)=(a-b)+2(x-a)>a-b.综上,对任意实数x有|x-a|+|x-b|≥a-b,当且仅当b≤x≤a时等号成立.
CONCLUSION Let the real numbers a and b satisfy a≥b, then for any real number x, there is | xa | + | xb | ≥ab, and if and only if b≤x≤a, the equal sign holds. ) Since b, a are the zeros of | xb | and | xa | respectively, then we discuss in three cases: (1) | xa | + | xb | = (ax) + (2) when b≤x≤a, | xa | + | xb | = (ax) + (xb) = ab; (3) when x> a, | xa + xb) = (ab) +2 (xa)> ab. In summary, for any real number x we have | xa | + | xb | ≥ab, if and only if b≤x Equal sign is established when ≤a