|x|<1⇒0≤x<1 or -1<x≤0;
|y|<1⇒0≤y<1 or -1<y≤0.
We have 4 cases:
(1) 0≤x,y<1
(2) -1<x≤0, 0≤y<1
(3) 0≤x<1, -1<y≤0
(4) -1<x,y≤0
Consider z=|xy+1|-|x+y| for each of the 4 cases:
(1) z=xy+1-x-y=x(y-1)-(y-1)=(x-1)(y-1). Both factors are negative so their product is positive and |xy+1|>|x+y|.
(2) xy+1>0, x+y can be positive or negative.
(a) z=xy+1-x-y=(x-1)(y-1)>0 because x-1<0 when x<0; or
(b) z=xy+1+x+y=(x+1)(y+1)>0, because x>-1. Therefore |xy+1|>|x+y|.
(3) Since x and y are symmetrical in x and y, this case is similar to (2), therefore |xy+1|>|x+y|.
(4) xy+1>0 and |x+y|=-(x+y), z=xy+1-(-(x+y))=xy+1+x+y which is (2b), therefore |xy+1|>|x+y|.
In all cases |xy+1|>|x+y| QED