Find range of values of $a,b,c$ such that $ax^2+bx+c$ satisfies given conditions Let $y = f(x) = ax^2+bx+c$ where $a\neq0$. Find the range of values $a,b$ and $c$ such that it satisfies the following:

*

*$0 \le f(x) \le 1 \quad \forall \space x \in [0,1]$

What I have found so far?

*

*For $x=0$, we get $0 \le c \le 1$


*For $x=1$, we get $0 \le a+b+c \le 1$


*If roots exist, then they should lie outside the interval $[0,1]$.
Therefore, ${{-b - \sqrt{D}}\over{2a}} < 0$ and ${{-b + \sqrt{D}}\over{2a}} >1$. I tried to solve these but didn't get any fruitful results.
EDIT 1: As explained in comments by @insipidintegrator, this is only for distinct roots.


*Vertex of quadratic $f(x)$ is $\big( \frac{-b}{2a}, -\frac{D}{4a}\big)$. If the x-coordinate of the vertex lies between 0 and 1, then y-coordinate should also be between 0 and 1. But I have no idea how to proceed from here.
I want to find some lower/upper limits for $a,b,c$ or any kind of relation between them.  Any solution or hints for solving this problem is greatly appreciated.

We may obtain the necessary and sufficient conditions for
$$a, b, c \in \mathbb{R}, ~ 0 \le ax^2 + bx + c \le 1, \forall x \in [0, 1].$$
Fact 1: Let $A, B, C \in \mathbb{R}$. Then
$$Ax^2 + Bx + C \ge 0, ~ x \ge 0
\iff A \ge 0, ~ C \ge 0, ~ B \ge -\sqrt{4AC}.$$
(The proof is given at the end.)
Fact 2: Let $A, B, C \in \mathbb{R}$. Then
$$Ax^2 + Bx + C \ge 0, ~ x\in [0, 1] \iff 
\left\{\begin{array}{l}
 A + B + C \ge 0\\[5pt]
 C \ge 0\\[5pt]
 B + 2C \ge - \sqrt{4(A + B + C)C}.
\end{array}
\right.
$$
(The proof is given at the end.)
Now, using Fact 2, we have
$$ax^2 + bx + c \ge 0, ~ x \in [0, 1]
\iff 
\left\{\begin{array}{l}
 a + b + c \ge 0\\[5pt]
 c \ge 0\\[5pt]
 b + 2c \ge - \sqrt{4(a + b + c)c}
\end{array}
\right.$$
and
$$ - ax^2 - bx - c + 1 \ge 0,~ x \in [0, 1]$$
$$\iff 
\left\{\begin{array}{l}
 -a - b - c + 1 \ge 0\\[5pt]
 - c + 1 \ge 0\\[5pt]
 - b + 2(-c + 1) \ge - \sqrt{4(-a - b - c + 1)(-c + 1)}.
\end{array}
\right.$$
Thus, we have
$$0 \le ax^2 + bx + c \le 1, ~ x\in [0, 1]$$
$$\iff \left\{\begin{array}{l}
 0 \le a + b + c \le 1\\[5pt]
 0 \le c \le 1\\[5pt]
 b + 2c \ge - \sqrt{4(a + b + c)c}\\[5pt]
 b + 2c \le 2 + \sqrt{4(-a - b - c + 1)(-c + 1)}.
\end{array}
\right.$$
$\phantom{2}$

Proof of Fact 1:
“$\Longleftarrow$”: $\quad$ For $x\ge 0$, we have $Ax^2+Bx+C\ge Ax^2 - \sqrt{4AC}\ x + C = (x\sqrt{A} - \sqrt{C})^2\ge 0$.
“$\Longrightarrow$”: $\quad$ Clearly, $A \ge 0$ and $C \ge 0$.
If $B < - \sqrt{4AC}$ and $AC > 0$,
then $Ax_0^2 + Bx_0 + C < 0$ where $x_0 = \sqrt{\frac{C}{A}} > 0$. Contradiction.
If $B < - \sqrt{4AC}$ and $AC = 0$,
there exists $x_1 > 0$ such that $Ax_1^2 + Bx_1 + C < 0$ (easy). Contradiction.
We are done.
$\phantom{2}$
Proof of Fact 2:
By Fact 1, it suffices to prove that
$$Ax^2 + Bx + C \ge 0, ~ x\in [0, 1]
\iff (A + B + C)s^2 + (B + 2C)s + C \ge 0, ~ s \ge 0.$$
“$\Longleftarrow$”: $\quad$ Clearly, $A+B+C \ge 0$.
If $x = 1$, then $Ax^2 + Bx + C = A + B + C \ge 0$.
If $x\in [0, 1)$, letting $s = \frac{x}{1 - x} \ge 0$, we have
$x = \frac{s}{1+s}$ and
$$Ax^2 + Bx + C = \frac{1}{(1+s)^2}[(A+B+C)s^2+(B+2C)s+C] \ge 0.$$
“$\Longrightarrow$”: $\quad$ For $s\ge 0$, letting $x = \frac{s}{1+s}\in [0,1]$, we have
$$Ax^2+Bx+C = \frac{1}{(1+s)^2}[(A+B+C)s^2+(B+2C)s+C] \ge 0$$
which results in
$(A+B+C)s^2+(B+2C)s+C\ge 0$.
We are done.
