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?
 A: 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!
A: We can bound the sum of roots and improve the bound to $b\le\frac{-2\phi^{2}-2\phi-1}{4\phi\left(\phi+2\right)}=−0.405$, without needing Vieta's formula.
By the quadratic formula, the roots of $f\circ f(x)=f^{2}+af+b$ are $\underbrace{f(x)}_{x^{2}+ax+b}=\frac{-a\pm\sqrt{a^2-4b}}{2}$. Then, by applying the formula again, the roots of $f\circ f(x)$ must be
$$x=\frac{-a\pm\sqrt{a^{2}-2a-4b\pm2\sqrt{a^{2}-4b}}}{2}\tag{*}$$
As $f\circ f$ is a composition of two quadratics, it - and by extension, its roots - are symmetric across the vertical line through its minimum, $x=\frac{-a}{2}$. So, $(*)$ can be considered as specifying the roots' $4$ distances from a centre, $\frac{-a}{2}$. It is given that these four roots exist (possibly up to multiplicity) and they are indexed by the $\pm$ signs in $(*)$. I will attempt to justify their order. As it is given that $a^2-4b\ge0$, we have $\sqrt{a^2-4b}\ge0$, so
$$\begin{align}\color{red}{-}\sqrt{a^{2}-4b}&\le\color{red}{+}\sqrt{a^{2}-4b}
\\
\frac{\sqrt{a^{2}-4b-2a\color{red}{-}2\sqrt{a^{2}-4b}}}2&\le \frac{\sqrt{a^{2}-4b-2a\color{red}{+}2\sqrt{a^{2}-4b}}}2\tag{**}\end{align}$$
The last LHS radical represents the distance from $\frac{-a}2$ of the two roots with an inner $\pm$ sign of $-$, so they must be closer to the centre as this is the smaller of the two. Conversely, the last RHS represents an inner $\pm$ sign of $+$, and these roots must be further away. 
The argument of the outermost radical in $(**)$ is also nonnegative, so the order of the roots must be:
$$\frac{-a-\sqrt{+}}{2}\le \frac{-a-\sqrt{-}}{2}\le \frac{-a+\sqrt{-}}{2}\le \frac{-a+\sqrt{+}}{2}$$
Therefore, the sum of the roots of $f\circ f$ must be at least the sum of the two most negative roots and at most the sum of the two most positive roots. And it is given that the sum of two roots, $-1$ is within these bounds. The problem is therefore to show that the region of the $(a,b)$ plane that satisfies 
$$\left(\frac{-a-\sqrt{+}}{2}+ \frac{-a-\sqrt{-}}{2}\right)\le -1 \le \left(\frac{-a+\sqrt{-}}{2}+ \frac{-a+\sqrt{+}}{2}\right)$$
is a subset of the region $b\leq \frac{-1}{4}$. The curves that bound the regions are shown below and are those where the inequality is an equality and those where the radicals arguments' are zero: $b=\frac{a^{2}}{4},\frac{a^{2}-2a-2\pm2\sqrt{2a+1}}{4}$.

For the lower bound, we start with $
\frac{-a-\sqrt{a^{2}-4b-2a-2\sqrt{a^{2}-4b}}}{2}+\frac{-a-\sqrt{a^{2}-4b-2a+2\sqrt{a^{2}-4b}}}{2}\le-1
$. With the substitution $c=a^2-4b$ and evaluating the case when both sides are equal, we have $2a-2+\sqrt{c-2a+2\sqrt{c}}+\sqrt{c-2a-2\sqrt{c}}$, which solves for $b=\frac{-2a^{2}+2a-1}{4a\left(a-2\right)}$.  For the upper bound, the curve's boundary line can be treated similarly and solves for the same function, $b=\frac{-2a^{2}+2a-1}{4a\left(a-2\right)}$. When I tried solving this, it was algebraically inefficient and required the aid of a CAS. However, judging by the neat solution, it could probably be streamlined. From here, it is the trivial matter of showing that the permissible region is that below the minimum of $\frac{-2a^{2}+2a-1}{4a\left(a-2\right)}$ and $\frac{a^{2}-2a-2\pm2\sqrt{2a+1}}{4}$, where they are each defined, and that these curves have a maximum at $(-\phi,\frac{-2\phi^{2}-2\phi-1}{4\phi\left(\phi+2\right)})=(-1.62,−0.405)$ (red point on figure), which is indeed less than $-0.25$ (red line on figure). 
What is even more interesting is that we can use this same method of bounding the plane to show that if the sum of two roots was a constant, $c$, then $b\le -\frac{\left(c+1\right)}{2}a$, of which we were given the degenerate case $c=-1$.
