Iranian Mathematical Olympiad

$$\left|\frac{a}{b}-1\right|\le \frac{1}{2},\left|\frac{b}{c}-1\right|\le \frac{1}{2},\left|\frac{c}{d}-1\right|\le \frac{1}{2},\left|\frac{d}{a}-1\right|\le \frac{1}{2}.$$ 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 $$\begin{gathered} \left|\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right|=\left|\sum \frac{u-iv}{u^2+v^2}\right|\ge \left|\Re \left(\sum \frac{u-iv}{u^2+v^2}\right)\right| \\ =\left|\sum \frac{u}{u^2+v^2}\right|\ge \sum _{cyc}\frac{u}{u^2+v^2}. \end{gathered}$$

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.