this is a question in my textbook that I was working on, it's included in the chapter on complex-numbers .(I'm in highschool)
I have already demonstrated this proposition to be true using the proof by contradiction , by assuming its negation and deriving a contradiction at the end ;
$$\bbox[pink,5px] {(\forall z\in \mathbb{C}) : |1+z|\geqslant \frac{1}{2}\text{ or } |1+z^2|\geqslant 1} $$
but I'd like to know if there is another demonstation without using the proof by contradiction and directly by algebraic manipulation . Can u suggest another way to proving it ? Thanks in advance.
P.S : I can post my demonstration in the comments if it's needed .
Geometric proof: Suppose $|1 + z| < \frac 1 2$. Then $z$ lies in a disk of radius $1/2$ centered at $-1$; in particular, its argument lies between $3\pi/4$ and $5\pi/4$ (draw the picture for this!!).
Squaring $z$ doubles the argument, and so $z^2$ lies in the right half-plane and $|z^2 + 1| \ge 1$ (since the distance from $z^2$ to $-1$ is at least $1$). The claim follows.