MBA-day33 绝对值的几何意义

三角不等式应用

|a|-|b| <= |a+b| <= |a|+|b| 与 |a|-|b| <= |a-b| <= |a|+|b|

ab同号，即ab >= 0, 则：
|a+b| = |a| + |b|
|a-b| = |a| - |b|

ab异号，即ab <= 0, 则：
|a-b| = |a| + |b|
|a+b| = |a| - |b|

已知a, b为实数，则|a| <= 1, |b| <= 1

1. |a + b| <= 1
2. |a - b| <= 1
   答：1）2）单独都无法推出|a| <= 1, |b| <= 1, 1)和2)联合可以推出|a| <= 1, |b| <= 1

解：
由三角不等式应用 |x|-|y| <= |x+y| <= |x|+|y| 与 |x|-|y| <= |x-y| <= |x|+|y|

|a + b| <= 1，令 a + b = x,即 |x| < =1
|a - b| <= 1，令 a - b = y,即 |y| < =1

-> |a + b|-|a - b| <= |a + b + a - b| <= |a + b|+|a - b|
= |2a| <= 2
= |a| <= 1

-> |a + b|-|a - b| <= |a + b - (a - b)| <= |a + b|+|a - b|
=|2b| <= 2
=|b| <= 1

即1）2）单独都无法推出|a| <= 1, |b| <= 1, 1)和2)联合可以推出|a| <= 1, |b| <= 1