Modulus of complex number less then equal to 1

Question

It $|z+1|\le 1 \text{ and } |z^2+1|\le 1$, then we have $$ |z|\le 1.$$ I wrote $z=x+iy, x,y\in \mathbb{R}$ and the inequalities from hypothesis become \begin{equation} (x+1)^2+y^2\le 1 \text{ and } (x^2-y^2+1)^2+4x^2y^2\le 1 \end{equation}...and I don't see how to deduce from here $$x^2+y^2\le 1.$$ I represent with Wolfram the region and I obteined something that didn't help my intuition... first inequality, second one

Answer 1

Applying Triangle Inequality: $$2|z| = |z^2+2z+1 - 1 - z^2|\le |z+1|^2 + |z^2+1| \le 2$$