A family of parabolas $p(x)$ is given for $x \in [0,1]$
by coefficients $(a,b,c)$ , everything real-valued:
$$
p(x) = a x^2 + b x + c
$$
The area of the parabolas is normed: $\int_0^1 p(x)\, dx = 1$ ;
they have a minimum: $a > 0$ ; and they do not intersect the x-axis: $b^2 - 4ac < 0$ .
So the maximum values must be at $x=0$ or $x=1$ .
What is the upper bound of these maximum values ?
The answer IMO is not at all quite trivial. Using the Ansatz by mookid [ $p(x) = a\left( (x-m)^2 + M^2
\right)$ ]we find:
$$
\frac{a}{3} \left[(1-m)^3 + m^3\right] + aM^2 = 1
\quad \Longrightarrow \quad a = \frac{1}{1/3 - m + m^2 + M^2}
\quad \Longrightarrow \\ p(0)(m,M) = \frac{m^2+M^2}{1/3 - m + m^2 + M^2}
$$
As a function of $M$ , $p(0)(m,M)$ is maximal (or minimal) for $M=0$ :
$$
\frac{\partial\, p(0)(m,M)}{\partial M} = 0 \quad \Longrightarrow \quad M = 0
$$
Hence proceeding with $M=0$ :
$$
p(0)(m,0) = \frac{m^2}{1/3 - m + m^2} \\
\frac{dp(0)(m,0)}{dm} = 0 \quad \Longrightarrow \quad m \in \{ 0,2/3 \}
\quad \Longrightarrow \quad p(0)(2/3,0) = 4
$$
So the upper bound is $\huge{\;4\;}$ , which means that I can pinpoint
my graphics window at that value.
Notes.
$$
\frac{\partial\,p(0)(m,M)}{\partial M} =
\frac{2M(m-1/3)}{\left[M^2-(m-1/3)+m^2\right]^2} = 0
$$
It follows that $M=0$ is a solution, giving $\;p(0)(m,0)=m^2/(1/3-m+m^2)\;$ ,
leading to the upper bound as required. But there is also a solution $\;m=1/3\;$ ,
leading to a function value which is independent of the parabola's minimum $M$ : $\;p(0)(1/3,M) = 1$ .
The other way around:
$$
\frac{\partial\,p(0)(m,M)}{\partial m} =
\frac{M^2-m^2+2m/3}{\left[M^2-(m-1/3)+m^2\right]^2} = 0
$$
It follows that $m=1/3+\sqrt{1/9+M^2}$ is the solution.
Substitution in $\;p(0)(m,M)\;$ yields:
$$
p(0)(1/3+\sqrt{1/9+M^2},M) = 2\frac{1+\sqrt{1+9M^2}+9M^2}{18M^2-\sqrt{1+9M^2}+2}
$$
And the maximum is found for derivative zero:
$$
\frac{dp}{dM} = -\frac{54M\sqrt{1+9M^2}}{\left(18M^2-\sqrt{1+9M^2}+2\right)^2} = 0
\quad \Longrightarrow \quad M=0 \quad \Longrightarrow \quad m = 2/3
$$