First solution. Swapping, if necessary, the roles of $a$ and $c$ and those of $b$ and $d$, we may assume that $a \le c$. In this case, the function $f(x) = \frac{a+x}{c+x}$ is increasing on $[1, 2]$, while $g(x) = \frac{c+x}{a+x}$ is decreasing on $[1, 2]$, hence $\frac{a+b}{b+c} + \frac{c+d}{d+a} \le \frac{a+2}{c+2} + \frac{c+1}{a+1}$. On the other hand, $4 \cdot \frac{a+c}{b+d} \ge a+c$, therefore it is sufficient to prove that $\frac{a+2}{c+2} + \frac{c+1}{a+1} \le a+c$, which reduces to $a+c+4 \le a^2c+ac^2+a^2+3ac$. From $(a-1)(c-1) \ge 0$ and $a^2c+ac^2+a^2+2ac \ge 5$ the conclusion follows, with equality if $a=c=1$, and $b=d=2$.

Second solution. We show that
$$\frac{a+b}{b+c} \le 2 \cdot \frac{a+c}{b+2} \le 2 \cdot \frac{a+c}{b+d} \quad \text{and} \quad \frac{c+d}{d+a} \le 2 \cdot \frac{a+c}{b+d},$$
inequalities which give the conclusion. The second inequality can be obtained from the first one by swapping $a$ with $c$ and $b$ with $d$, therefore it is sufficient to prove the first one. This inequality reduces to $b^2 + ab + 2b + 2a \le 2c^2 + 2ac + 2bc + 2ab$. But $2a \le 2ac$, $2b \le 2bc$ and $b^2 \le b + 2 \le ab + 2c^2$, the inequality $b^2 \le b + 2$ being equivalent to $(b+1)(b-2) \le 0$. We obtain the equality case $a=1, b=2, c=1$ and $d=2$.

Third solution. As $0 < b+d \le 4$, it is sufficient to prove that the left hand side is at most $a+c$. Without loss of generality, we may assume that $a \le c$. In this case,
$$\frac{a+b}{b+c} + \frac{c+d}{d+a} \le 1 + \left(1 + \frac{c-a}{d+a}\right) \le 2 + c-a \le c+a,$$
with equality if $a=1, b=2, c=1$ and $d=2$.