证明|x-a|+|x-b|≥|a-b|和|x-a|-|x-b|≤|a-b|

Question

证明|x-a|+|x-b|≥|a-b|和|x-a|-|x-b|≤|a-b|

Answer 1

首先,证明绝对值不等式
|a+b|² = a²+b²+2ab≤|a|²+|b|²+2|a||b| = (|a|+|b|)²,当且仅当ab≥0时等号成立
|a+b|² = a²+b²+2ab≥|a|²+|b|²-2|a||b| = (|a|-|b|)²,当且仅当ab≤0时等号成立
所以|a|-|b| ≤ |a+b| ≤ |a|+|b|
其次,根据上述绝对值不等式
|x-a|+|x-b| = |a-x|+|x-b| ≥ |(a-x)+(x-b)| = |a-b|
|x-a|-|x-b| = |a-x| - |x-b|≤ |(a-x)+(x-b)| = |a-b|