Iranian Mathematical Olympiad

$$\left| \frac{a}{b} - 1 \right| \le \frac{1}{2}, \quad \left| \frac{b}{c} - 1 \right| \le \frac{1}{2}, \quad \left| \frac{c}{d} - 1 \right| \le \frac{1}{2}, \quad \left| \frac{d}{a} - 1 \right| \le \frac{1}{2}.$$
Putting $(\frac{a}{b}, \frac{b}{c}, \frac{c}{d}, \frac{d}{a}) = (x, y, z, t)$ such that $|x-1| \le \frac{1}{2}$, $|y-1| \le \frac{1}{2}$, $|z-1| \le \frac{1}{2}$, $|t-1| \le \frac{1}{2}$ we are going to prove that
$$\left| \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{t} \right| > \frac{7}{2}.$$
by letting $x = u + iv$, for some real numbers $u$, $v$ such that $u^2 + v^2 \le 2u - \frac{3}{4}$. It follows that $u \ge \frac{3}{8}$. Since
$$\frac{1}{x} = \frac{u - iv}{u^2 + v^2},$$
We would obtain
$$\left| \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{t} \right| = \left| \sum \frac{u - iv}{u^2 + v^2} \right| \ge \left| \Re \left( \sum \frac{u - iv}{u^2 + v^2} \right) \right| \\ = \left| \sum \frac{u}{u^2 + v^2} \right| \ge \sum_{cyc} \frac{u}{u^2 + v^2}.$$

Since $u \ge \frac{1}{2}(u^2 + v^2 + \frac{3}{4})$ we would obtain
$$\sum_{cyc} \frac{u}{u^2 + v^2} \ge \frac{1}{2} \sum_{cyc} \frac{u^2 + v^2 + \frac{3}{4}}{u^2 + v^2} = 2 + \frac{3}{8} \sum_{cyc} \frac{1}{u^2 + v^2} = 2 + \frac{3}{8} \sum_{cyc} \frac{1}{|x|^2}.$$
Notice that $xyzt = 1$ thus, according to AM-GM inequality,
$$\sum_{cyc} \frac{1}{|x|^2} \ge 4 \sqrt[4]{\frac{1}{|x|^2 |y|^2 |z|^2 |t|^2}} = 4.$$
Hence,
$$\sum_{cyc} \frac{u}{u^2 + v^2} \ge 2 + 4 \cdot \frac{3}{8} = \frac{7}{2}.$$
Notice that this inequality has no equality case.