How I do evaluate this polynomial problem? Let $f(x)=x^2+ax+b$.
Suppose $f(f(x))=0$ equation has $4$ different real solutions $x_1,x_2,x_3,x_4$
and that two of them sum up to $-1$ (i.e. $\exists\, i\neq j$ such that $x_i+x_j=-1$)
Prove that $b\lt-\frac14$

Let $y = f(x)$ then we have the following system
$$ \begin{cases} x^2 + ax + b = y & (1) \\ y^2 + ay + b = 0 & (2) \end{cases} $$
In order to have 4 distinct real solutions in $x$, we require two distinct real solutions in $y$, therefore $a^2-4b > 0$. Denote the two solutions as $y_1$ and $y_2$, then
$$ \begin{cases} x^2 + ax + b = y_1 & (3) \\ x^2 + ax + b = y_2 & (4) \end{cases} $$
We now need two dinstinct solutions for each value of $y$, therefore another necessary condition is $a^2 - 4(b-y_i) > 0$ or $a^2 - 4b + 4y_i > 0$, where $i=1,2$. Then $$ 2(a^2-4b) + 4(y_1+y_2) = 2(a^2-4b) - 4a > 0 $$
or $$4b < a^2 -2a \tag{5}$$
Denote $x_1,x_2$ as solutions to $(3)$ and $x_3,x_4$ as solutions to $(4)$. Then it follows that $$x_1 + x_2 = x_3 + x_4 = -a \tag{6} $$
If $a=1$, then $(6)$ satisfies the condition and $(5)$ gives $b < -\frac{1}{4}$. 

If $a\ne 1$, it must be true that one root from $(3)$ and one root from $(4)$ add up to $-1$. Suppose $x_1 + x_3 = -1$, then adding $(3)$ and $(4)$ gives
$$ {x_1}^2 + {x_3}^2 + a(x_1+x_3) + 2b = y_1 + y_2 $$
$$ {x_1}^2 + {x_3}^2 = - 2b $$
Then
$$ 2x_1x_3 = (x_1+x_3)^2 - ({x_1}^2 + {x_3}^2) = 1 + 2b $$
It follows that $x_1$ and $x_3$ must be roots of the equation
$$ x^2 + x + \frac{1+2b}{2} = 0 $$
Since $x_1 \ne x_3$, we require
$$ 1 - 2(1+2b) > 0 $$
or $b < -\frac{1}{4}$
