Chinese Girls' Mathematical Olympiad

Notice that when $\sqrt{ac} \le x$, we have
$$\frac{1}{x+a} + \frac{1}{x+c} - \frac{2}{x+\sqrt{ac}} = \frac{(\sqrt{a}-\sqrt{c})^2(\sqrt{ac}-x)}{(x+a)(x+c)(x+\sqrt{ac})} \le 0. \quad (*)$$
Given the conditions, $\sqrt{ac} \le 1 \le 1+b$, and $\sqrt{ac} \le 1+d$. Substituting $x = 1+b$ and $x = 1+c$ into $(*)$ yields
$$\frac{1}{1+a+b} + \frac{1}{1+b+c} \le \frac{2}{1+b+\sqrt{ac}},$$
$$\frac{1}{1+c+d} + \frac{1}{1+d+a} \le \frac{2}{1+d+\sqrt{ac}}.$$
This means that when substituting $\sqrt{ac}$ for $a$ and $c$, the left side of the inequality does not decrease, while the right side remains unchanged. Thus, we may assume without loss of generality that $a = c$. Similarly, we may assume $b = d$. Hence, the original inequality is reduced to proving
$$\frac{1}{1+a+b} \le \frac{1}{1+2\sqrt{ab}}.$$
This holds true by the arithmetic mean-geometric mean inequality, which states $a+b \ge 2\sqrt{ab}$. $\square$