Let $z\in C$, where $C$ is the unit square in the complex plane with diagonal corners at $0, 1, i, 1+i$.
From a geometrical point of view, it is clear to me that since $(5,0)$ is closest to the corner at $(1,0)$, we must have $|\bar{z}-5|\geq4$.
I'm trying to show it algebraically, but using the reverse triangle inequality I have only managed to show that $|\bar{z}-5|\geq5-\sqrt{2}$, which is not enough.
Any suggestions?
Let $z=a+bi$. Show that if $|a|,|b|<1$ then $(5-a)^2+b^2\geq 16$.
Now, $(5-a)^2 + b^2\geq (5-a)^2$ and $5-a\geq 4$. So....