Verified answer

Note that g(x+y) ≤ g(x)+g(y)

<==> |x+y|/(1+|x+y|) ≤ |x|/(1+|x|) + |y|/(1+|y|)

<==> 1 - 1/(1+|x+y|) ≤ (1 - 1/(1+|x|)) + (1 - 1/(1+|y|))

<==> 1/(1+|x|) + 1/(1+|y|) ≤ 1 + 1/(1+|x+y|)

<==> [(1+|x|) + (1+|y|)] / [(1+|x|)(1+|y|)] ≤ [(1+|x+y|) + 1)/(1+|x+y|)

<==> (2 + |x| + |y|) / [(1+|x|) (1+|y|)] ≤ (2 + |x+y|) / (1+|x+y|)

<==> (2 + |x| + |y|) (1 + |x+y|) ≤ (2 + |x+y|) (1 + |x|) (1 + |y|)

<==> 2 + |x| + |y| + 2|x+y| + |x||x+y| + |y||x+y|

≤ 2 + 2|x| + 2|y| + 2|xy| + |x+y| + |x||x+y| + |y||x+y| + |xy||x+y|

<==> |x+y| ≤ |x| + |y| + 2|xy| + |xy||x+y|.

This is true, since |x+y| ≤ |x| + |y| by the triangle inequality, and so

|x+y| ≤ |x| + |y| ≤ |x| + |y| + 2|xy| + |xy||x+y|, since 2|xy| + |xy||x+y| ≥ 0.

I hope this helps!