Given: A,B∈ℝ, |A|<∞.
|B||A|≥0 because |B| and |A| are both positive by definition of the absolute function (||).
If |B|<|A|, then |B|-|A|<0, hence |B||A|≥|B|-|A|, because all positive quantities are greater than negative quantities.
If |B|=|A| then |B|-|A|=0, so |B||A|≥|B|-|A| because |B||A|≥|B|-|A|=0.
If |B|>|A| then |B|=|A|+k where k>0; |B||A|=(|A|+k)|A|=A2+k|A|; |B|-|A|=k.
|B||A|-(|B|-|A|)=A2+k|A|-k=A2+k(|A|-1). If |A|<1 then A2<|A| so k(|A|-1)<0 and A2+k(|A|-1) can be negative.
For example, let |A|=½, then A2=¼, so we have ¼-½k<0 when ¼<½k, that is, when k>½.
This means that if, for example, |A|=½ and |B|=2, then |B||A|-(|B|-|A|)=1-1½<0, so |B||A|<|B|-|A|.
Therefore the proposed statement is false.