<s>[INST] 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 $b\lt-\frac14$
 [/INST] $f(x_k)$ 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.
</s>