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 $b\lt-\frac14$
 A: I assume that $x_1,x_2$ are the roots of $f(x)=0$ ?
Since $f(x)=x(x+1)$ has $b=0$ and that $f\circ f(x)=0$ has four solutions $0,-1,j,j^2$.
I will also assume that you want four real solutions to the equation.

Let's rewrite $f(x)=(x-t)(x+t+1)$ with $x_1=t$ and $x_2=-1-x_1$
We have $b=-t(t+1)$
And $f(f(x))=(-x^2-x+t^2+2t)(-x^2-x+t^2-1)$
Which has solutions $\begin{cases}-\frac 12\pm\frac 12\sqrt{4t^2-3}\\-\frac 12\pm\frac 12\sqrt{4t^2+8t+1}\end{cases}$
The condition of positivity of the inners of the square roots results in $$t\in\left]-\infty,-1-\frac{\sqrt{3}}2\right[\cup\left]\frac{\sqrt{3}}2,+\infty\right[$$
Conditionning $b<-\dfrac{2\sqrt{3}+3}4\approx -1.616$

Edit: assuming new condition that $x_1,x_2$ are two roots of $f(f(x))=0$ whose sum is $-1$.

Let's write $f(x)=(x-u)(x-v)$ with $u+v=-a$ and $uv=b$
Then $f(f(x))=\left(x^2-(u+v)x+uv-u\right)\left(x^2-(u+v)x+uv-v\right)$
Which has solutions $\begin{cases}\frac {u+v}2\pm\frac 12\sqrt{(u-v)^2+4u} & \text{let's call them }x_u,\bar x_u &\text{with radical }\delta_u\\\frac {u+v}2\pm\frac 12\sqrt{(u-v)^2+4v} & \text{let's call them }x_v,\bar x_v &\text{with radical }\delta_v\end{cases}$
When $u\neq v$ the solutions provided they exists are then all different.
The condition that the sum of two roots is $-1$


*

*If we take roots with the same radical then $x_u+\bar x_u=x_v+\bar x_v=u+v=-1$ and we are back to the former problem. 

*Thus let's examine a crossed sum with different radicals $u+v+\frac 12(\pm\delta_u\pm\delta_v)=-1$.
To be continued...
A: $f(x_k)$ are real roots of $f$, not all the same, so that $4b<-a^2$ is a necessary condition. If $z=f(x_2)=f(x_2)$ are the same root of $f$, then $x_1,x_2$ are the two solutions of $$f(x)=z\iff 0=x^2+ax+b-z$$ so that by Viete $a=-(x_1+x_2)=1$ and $b<-\frac14$ is necessary.
If $f(x_1)\ne f(x_2)$ then by Viete 
\begin{align}
-a&=f(x_1)+f(x_2)=x_1^2+x_2^2-a+2b\\
\iff x_1^2+x_2^2&=-2b,\qquad 1+2b = 2x_1x_2\\
b&=f(x_1)f(x_2)\\
&=x_1^2x_2^2+ax_1x_2(x_1+x_2)+b(x_1^2+x_2^2)+ab(x_1+x_2)+b^2  \\
&=\tfrac14+b+b^2-a(\tfrac12+b)-2b^2-ab+b^2\\
\iff \tfrac14&=a(\tfrac12+2b)
\end{align}
so that in combination with the necessary condition
$$
\frac1{2a}-1=4b<-a^2
$$
If $b<-\frac14$ were not satisfied, $a$ would have to be positive. But then by the arithmetc-geometric mean inequality
$$
1>\frac1{2a}+a^2=3\,\frac{\frac1{4a}+\frac1{4a}+a^2}3\ge 3\,\sqrt[3\;]{\frac1{16}}=\sqrt[3\;]{\frac{27}{16}}>1
$$
which is impossible.
A: 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 $b < -\frac{1}{4}$
