Iranian Mathematical Olympiad

a) Let $O$ be the origin point of the complex plane. Consider numbers $x, y$ as points on this plane. The given inequality implies
$$\angle yOx \leq \frac{\pi}{3}.$$
But since $Oxy$ is an isosceles triangle with $|Ox| = |Oy| = 1$, this means $|xy| = |x - y| \geq 1$. Therefore for any complex number $z$ it is deduced that
$$\begin{aligned} & |z - x| + |z - y| \geq |x - y| \geq 1 \\ \implies & |z| + |z - x| + |z - y| \geq |z| + 1 = |zx| + |y| \geq |zx - y| \end{aligned}$$

b) The given inequality implies triangle $Oxy$ has three angles, all less than or equal to $\frac{2\pi}{3}$, so its first Fermat point does not lie outside of it. The expression $|z| + |z - x| + |z - y|$ is the sum of distances from point $z$ to the vertices of $\triangle Oxy$. This sum is minimized when $z$ is the Fermat point. So it's needed to calculate the given sum for the Fermat point. Let $p$ be the clockwise rotation of $x$ with respect to the origin point and angle $\frac{\pi}{3}$, i.e. $p = \text{cis}(\frac{\pi}{3})x = \left|\frac{1}{2} + \frac{\sqrt{3}}{2}i\right| x$. Due to the properties of Fermat point, the desired value is the distance between $y$ and $p$, which is
$$|y - p| = |yi - pi| = \left|yi - \left(\frac{1}{2}i - \frac{\sqrt{3}}{2}\right)x\right| = \left|\frac{\sqrt{3}}{2}x + \left(y - \frac{1}{2}\right)i\right|,$$
hence the desired inequality holds.