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 $$\angle yOx\le \frac{\pi }{3}.$$ But since $Oxy$ is an isosceles triangle with $|Ox|=|Oy|=1$, this means $|xy|=|x-y|\ge 1$. Therefore for any complex number $z$ it is deduced that $$\begin{array}{rl}& |z-x|+|z-y|\ge |x-y|\ge 1\\ ⟹& |z|+|z-x|+|z-y|\ge |z|+1=|zx|+|y|\ge |zx-y|\end{array}$$

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 $△Oxy$. 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.