Q:

Complex numbers inequality done geometricaly?

Let $a \in \mathbb{C}$. Prove that, if $|a+z| \leq |a+1|, \: \forall z \in \mathbb{C}$ such that $|z|=1$, then $a \in [0, \infty)$.
After squaring I got $z\overline{a}+\overline{z}a \leq a+ \overline{a}$ and after letting $a=m+ni$ and $z=p+qi$ with $p^2+q^2=1$, I obtained $m \geq 0$ and $n \leq m\frac{\sqrt{1-p}}{\sqrt{1+p}}$, for $p \in [-1,1)$. Choosing $p \in [0,1)$ we can easily see that $n \leq 0$. Similarly we can get $n\geq 0$, thus $m \geq 0$ and $n=0$, so $a \in [0,\infty)$.
What I'm asking is whether there is a nice geometrical solution or not. Thanks!

A:

Here's a geometric interpretation and an alternative proof. Consider the number $a$ and the disk on $\mathbb{C}$ given by all the numbers of absolute value less or equal than $|a+1|$. Then, if $a$ is not pointing to the same direction as $1$, we can "move along" $a$ for one unit distance and end up outside the disk of radius $|a+1|$. Drawing a picture with $a$, $a+1$, disk of radius $|a+1|$ and $a+a/|a|$ makes this intuition clear. 
Algebraically, the same idea gives another proof of the statement: let's pick $z = a/|a|$, so $|z| = 1$ and see what happens: 
$$
|a+a/|a||= |a(1+1/|a|)| = |a|(1+1/|a|) = |a| + 1
$$
This choice of $z$ actually maximizes $|a+z|$, which at least geometrically should be clear. Now by the triangle inequality
$$
|a+1| \leq |a|+1.
$$
In particular the inequality is strict when $a \notin [0, \infty)$, which can be seen by squaring both sides (or just geometrically thinking about the triangle with sides of length $|a+1|$, $|a|$ and $1$). 
So $|a+1| \leq |a+a/|a||$ with strict inequality whenever $a$ is not a non-negative real number.

