Suppose that $z\in\mathbb C$ with $|z^2+1|\le 1$. How to prove $|z+1|\geq\frac12$. Let $z\in\mathbb C$ with $|z^2+1|\le 1$. I want to prove $|z+1|\geq\frac12$. 
I noticed that $|z^2+1|\le 1$ means that $z$ lies in a cassini oval. I tried with the substitution $z=r\exp(i\theta)$ where $r\geq0, \theta\in[0,2\pi]$. The constraint now is $\frac{r^2}2\le-\cos(2\theta)$ and we need to prove $$\sqrt{r^2 \sin^2(\theta) + (r \cos(\theta) + 1)^2}=\sqrt{r^2+2r\cos\theta+1}\geq\frac12.$$
How can I do that?

It suffices to prove that
$$|z+1| < \tfrac{1}{2} \quad \Longrightarrow \quad |z^2+1| > 1.$$
To this end, let $z = a + \mathrm{i}b$. 
From $|z+1| < \frac{1}{2}$, we have
$(a+1)^2 + b^2 < \frac{1}{4}$ which results in
$a^2 - b^2 > a^2 + (a+1)^2 - \frac{1}{4} = 2(a+\frac{1}{2})^2 + \frac{1}{4} > 0$.
Then, we have
$|z^2 + 1| \ge \mathrm{Re}(z^2+1) = a^2 - b^2 + 1 > 1$. We are done.
