Solution:
First, we will prove that $f(x) = 0$ has some solutions (maybe not distinct).
Indeed, if $f(x) = 0$ has no root, then it can be written as $f(x) = (x - c)^{2} + d$ with $d > 0$ and
$$f(f(x)) = \left((x - c)^{2} + d - c\right)^{2} + d > 0.$$
It means $f(f(x)) = 0$ has no solution, which is a contradiction.

Now, denote $c_{1} \geq c_{2}$ as the solution of $f(x) = 0$ and $x_{1}, x_{2}$ as the solutions of $f(f(x)) = 0$ in such a way that $x_{1} + x_{2} = -1$. By Vieta's theorem, note that $c_{1} + c_{2} = -a$ and $c_{1} c_{2} = b$.

It is easy to see that $f(f(x)) = 0$ is equivalent to $f(x) = c_{1}$, $f(x) = c_{2}$. We need to consider 2 cases:

1. If $x_{1}, x_{2}$ are the solutions of one equation, by Vieta's theorem, then $a = 1$. Thus $c_{2} \leq -\frac{1}{2}$.

Consider equation $f(x) - c_{2} = 0$, we have $\Delta = 1 - 4(b - c_{2}) > 0$, which implies that $b < -\frac{1}{4}$.

2. If $x_{1}, x_{2}$ are solutions of two equations, then $x_{i}^{2} + a x_{i} + b_{i} = c_{i}$ with $i = 1, 2$. Sum these two identities, we get
$$x_{1}^{2} + x_{2}^{2} - a + 2b = -a \Leftrightarrow x_{1}^{2} + x_{2}^{2} + 2b = 0.$$
Hence, $b = -\frac{x_{1}^{2} + x_{2}^{2}}{2} \leq -\frac{(x_{1} + x_{2})^{2}}{4} = -\frac{1}{4}$.

Therefore, in all case, we always have $b \leq -\frac{1}{4}$.