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?
 A: 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 $$\bbox[15px,border:1px groove navy]{|1+z|=\sqrt{\color{orange}{(1+a)^2+b^2}}\geq\sqrt{2-\sqrt 2}>\frac12.}$$
A: The curves that satisfy
$$|z^2+1| = 1,\>\>\>\>\>|z+1| = \frac12\tag 1$$
can be expressed in polar coordinates $z=re^{i\theta}$ as 
$$r^4+2r^2\cos2\theta=0, \>\>\>\>\> r^2+2r\cos\theta + \frac34=0 $$
Eliminate the variable $\theta$ to obtained the following quadratic equation in $r^2$,
$$2r^4-\frac12r^2+\frac9{16}=0$$
Its discriminate is negative, meaning the two curves given by (1) have no intersections. (See the plot below.) 
It is straightforward to verify that $z=-1$, the center of the circle $|z+1| = \frac12$, does not satisfies the $|z^2+1| \le 1$, which means the circle's center lies outside the region $|z^2+1| \le 1$. Then, along with the earlier conclusion that the circle does not intersect the other curve, we further conclude that the circle itself $|z+1| = \frac12$ is also outside the given region $|z^2+1| \le 1$. Thus,
$$|z+1|>\frac12$$

A: 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.
