(1) Let A=-a2 and B=b2, then A-B=-a2-b2=-(a2+b2), and |A-B|=a2+b2; |A|-|B|=a2-b2; a2+b2>a2-b2⇒|A-B|>|A|-|B|.
(2) Let A=a2 and B=-b2, then A-B=|A-B|=a2+b2; |A|-|B|=a2-b2; a2+b2>a2-b2⇒|A-B|>|A|-|B|.
(3) Let A=a2 and B=b2, then |A|-|B|=a2-b2, which can be positive or negative; but |A-B|=|a2-b2|>0 always. (3.1) When A<B, |A|-|B|<0, so |A-B|> |A|-|B|.
(3.2) When A>B, |A-B|=A-B; |A|-|B|=A-B, so |A-B|=|A|-|B|.
(4) Let A=-a2 and B=-b2, then |A-B|=|-a2+b2|; |A|-|B|=a2-b2, which can be positive or negative.
(4.1) When A>B, -a2>-b2, b2>a2. |A-B|=b2-a2>0; |A|-|B|=a2-b2<0, |A-B|>|A|-|B|.
(4.2) When A<B, -a2<-b2, b2<a2. |A-B|=a2-b2; |A|-|B|=a2-b2, |A-B|=|A|-|B|.
From these results the following conditions satisfy the inequality: (1), (2), (3.1), (4.1).
(i) A and B have opposite signs satisfies the inequality.
(ii) A and B are both positive then A<B satisfies the inequality.
(iii) A and B are both negative then A>B satisfies the inequality.