Iranian Mathematical Olympiad

First note that $(a+c)\left(\frac{1}{a}+\frac{1}{c}\right) = \frac{a}{c} + \frac{c}{a} + 2 = \left(\sqrt{\frac{a}{c}} + \sqrt{\frac{c}{a}}\right)^2$ and so
$$\sqrt{(a+c)\left(\frac{1}{a}+\frac{1}{c}\right)} = \sqrt{\frac{a}{c}} + \sqrt{\frac{c}{a}}$$
Similarly $\sqrt{(b+d)\left(\frac{1}{b}+\frac{1}{d}\right)} = \sqrt{\frac{b}{d}} + \sqrt{\frac{d}{b}}$.
On the other hand, for the left hand side of inequality, because of the hypothesis $\sum_{cyc} \frac{1}{ab} = 1$, we have
$$abcd = (a+c)(b+d)\left(\frac{1}{a}+\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{d}\right) = \left(\sqrt{\frac{a}{c}}+\sqrt{\frac{c}{a}}\right)^2\left(\sqrt{\frac{b}{d}}+\sqrt{\frac{d}{b}}\right)^2$$
So we must prove $X^2Y^2 + 16 \ge 8(X + Y)$, where $X = \sqrt{\frac{a}{c}} + \sqrt{\frac{c}{a}}$ and $Y = \sqrt{\frac{b}{d}} + \sqrt{\frac{d}{b}}$. Note that by AM-GM inequality $X \ge 2$ and $Y \ge 2$. Therefore, $(X - 1)(Y - 1) \ge 1$ and consequently, $XY \ge X + Y$. This implies the assertion, because $(XY)^2 + 16 \ge 8XY \ge 8(X + Y)$.