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?

Let $z=a+ib$, then $$|1+z^2|^2=1+2a^2+a^4+b^2(2a^2-2)+b^4\le1.$$ We want to minimize $|1+z|^2=\color{orange}{(1+a)^2+b^2}$. Let me complete the square in the constraint: $$b^4+b^2(2a^2-2)\color{blue}{+(a^2-1)^2}\le\color{blue}{(a^2-1)^2}-a^4-2a^2=1-4a^2$$ which is the same as $$(b^2+(a^2-1))^2\le1-4a^2.$$ In particular, it follows that $- \frac12<a<\frac12$ and $$b^2\in\left[1-a^2-\sqrt{1-4a^2},1-a^2+\sqrt{1-4a^2}\right].$$
Thus, $$2+2a-\sqrt{1-4a^2}\le\color{orange}{(1+a)^2+b^2}\le2+2a+\sqrt{1-4a^2}.$$ We will thus want to minimize the function $2a-\sqrt{1-4a^2}$. This can be done for example with calculus, but we can also proceed as follows:
Now we have $$8a^2-4 \sqrt 2 a+1=8 \left(a - \frac{1}{2 \sqrt 2}\right)^2\geq0$$ which implies $$4a^2-4\sqrt 2a+2=(\sqrt 2-2a)^2\geq1-4a^2$$ and thus $$\sqrt 2-2a\geq\sqrt{1-4a^2}.$$ By replacing $a$ by $-a$ we also get $$\sqrt{1-4a^2}\le\sqrt2+2a.$$
It follows from the previous result that $$2-\sqrt 2\le\color{orange}{(1+a)^2+b^2}\le2+\sqrt 2$$ and thus in particular $$\bbox[15px,border:1px groove navy]{|1+z|=\sqrt{\color{orange}{(1+a)^2+b^2}}\geq\sqrt{2-\sqrt 2}>\frac12.}$$
