We notice that
$$\frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+d} + \frac{d}{1+a} + abcd \le \frac{a}{1+abcd} + \frac{b}{1+abcd} + \frac{c}{1+abcd} + \frac{d}{1+abcd} + abcd = \frac{a+b+c+d}{1+abcd} + abcd.$$
Using repeatedly the inequality $x + y \le 1 + xy$, $\forall x, y \in [0, 1]$ (equivalent to $(1-x)(1-y) \ge 0$, therefore true), we get $a + b + c + d \le 1 + ab + 1 + cd = ab + cd + 2 \le 1 + abcd + 2 = abcd + 3$.

Replacing $x = abcd \in [0, 1]$, it is enough to prove that $1 + \frac{2}{1+x} + x \le 3$, that is $2 + x^2 + x \le 2x + 2$, or $x(1-x) \ge 0$, which is clearly true.