22nd Chinese Girls' Mathematical Olympiad

Notice that when $\sqrt{ac}\le x$, we have $$\frac{1}{x+a}+\frac{1}{x+c}-\frac{2}{x+\sqrt{ac}}=\frac{(\sqrt{a}-\sqrt{c}{)}^2(\sqrt{ac}-x)}{(x+a)(x+c)(x+\sqrt{ac})}\le 0.(\ast )$$ 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 $(\ast )$ 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 $a+b\ge 2\sqrt{ab}$. $\square$