For $f(x)=x^2+a\cdot x + b$ prove $b\leq -\frac{1}{4}$ Equation $f(f(x))=0$, for $f(x)=x^2+a\cdot x + b$, have four real solutions and sum of two of them is $-1$. Prove that $b\leq-\frac{1}{4}$.
My progress so far:
Let $x_{1}$ and $x_{2}$ be real solutions for $f(x)=0$. We know that discriminant of this equation is positive, so from $a^2-4b\geq 0$, we have $b<\frac{a^2}{4}$.
Now let $y_{1}$ and $y_{2}$ be real solutions for $f(x)=x_{1}$ and $y_{3}$ and $y_{4}$ be real solutions for $f(x)=x_{2}$. From Vieta's formulas we know $y_{1}+y_{2}=-a$ and $y_{3}+y_{4}=-a$. From here I don't know how to continue, had many ideas but none of them worked.
Can anyone help?

I found out a solution:
Discriminant of equations $f(x)=x_{1}$ and $f(x)=x_{2}$ must be positive so:
$a^2-4b+4x_{1}\ge 0$ and $a^2-4b+4x_{2}\ge 0$, if we sum them we will get:
$2a^2-4a-8b\ge 0$ (by vieta's formula $x_{1}+x_{2}=-a$)
form this we have $b\le \frac{1}{4}(a-1)^2-\frac{1}{4}$ ... (*)
Now discussion by which solutions sum to $-1$:
$1)$ $y_{1}+y_{2}=-1$ or $y_{3}+y_{4}=-1$ 
In this case by vieta's formula $y_{1}+y_{2}=-a=-1$ or $y_{3}+y_{4}=-a=-1$ and we have $a=1$ and by (*) $b\le -\frac{1}{4}$.
$2)$ $y_{1}+y_{3}=-1$ or any other pair of solutions
we have: $y_{1} = Y$, $y_{2}=-a-Y$, $y_{3}=-1-Y$ and $y_{4}=-a+1+Y$. From $x^2+ax+b-x_{1}$ we got $x_{1}=Y^2+aY+b$ and from $x^2+ax+b-x_{2}$ we got $x_{2}=Y^2-(a-2)Y+b-a+1$.
Now by $x_{1}+x_{2}=-a$ we have $Y^2+Y+b+\frac{1}{2}$. From problem we know that $Y$ is real , so discriminant must be positive or zero. That means $1-4(b+\frac{1}{2})\ge 0$, form that $b\le-\frac{1}{4}$.
Q.E.D.
This solution maybe isn't best or smarter but works!
