Is $\max\limits_{x \in [0,1]} |x-c| x(1-x)$ minimal if and only if $c=0.5$ Is the following statement true?
$$\max_{x \in [0,1]} |x-c| x(1-x) \text{ is minimal } \Leftrightarrow \max_{x \in [0,1]} |x-c|\text{ is minimal } \Leftrightarrow c=0.5$$
This question arises in some calculations related to an optimization problem. I reduced it to what is relevant. The second equivalence is clear, but I am not sure about the first one.
 A: We have to study the function
$$f_c(x):=(x-c)x(1-x)$$
on the $x$-interval $[0,1]$. Its graph is a cubical parabola with three real zeros $c$, $0$, $1$,  and highest coefficient $<0$. It follows that $f$ has  a local minimum and to the right of it a local maximum. Due to symmetry we may assume $c\leq{1\over2}$.
If $c<0$ the local minimum of $f_c$ is between $c$ and $0$, and the local maximum at $g(c)\in\ ]0,1[\ $, whereby $g(c)$ is the righthand zero of $f_c'$, hence
$$g(c)={1\over3}\bigl(1+c+\sqrt{1-c+c^2}\bigr)\ .$$
The value $f\bigl(g(c)\bigr)$ is then also the maximum of $|f_c|$ on $[0,1]$.
If $0<c\leq{1\over2}$ the local minimum of $f_c$ is between $0$ and $c$, and the local maximum of $f_c$ is between $c$ and $1$, whereby the upward bump of $f_c$ is higher than the downward bump, since $c\leq{1\over2}$. It follows that again the maximum of $|f_c|$ on $[0,1]$ is at $g(c)$.
It remains to study the function
$$\psi(c):=f_c\bigl(g(c)\bigr)\qquad\left(-\infty<c\leq{1\over2}\right)\ .$$
Unfortunately no simple expression for $\psi$ results. But we can show that  $\psi$ is decreasing, hence minimal at $c={1\over2}$.
Proof. Write$$F(x,c):=f_c(x)=(x-c)x(1-x)\ .$$
Then $\psi(c)=F\bigl(g(c),c\bigr)$, and therefore
$$\eqalign{\psi'(c)&=F_{.1}\bigl(g(c),c\bigr)g'(c)+F_{.2}\bigl(g(c),c\bigr)\cr  &=0+\bigl(-x(1-x)\bigr)_{x=g(c)}\cr  &=-g(c)\bigl(1-g(c)\bigr)<0\ .\qquad\square\cr}$$
A: Note that $$\hat{c}= \text{arg minimal}_c\{ \max_{x\in [0,1]} |x-c|x(1-x) \} \leq \text{arg minimal}_c\{ \max_{x\in [0,1]} [\frac{|x-c|+x+1-x}{3}]^3 \}(\text{Using AM-GM inequality)} = \text{arg minimal}_c\{ \max_{x\in [0,1]} [\frac{|x-c|+1}{3}]^3 \}\Leftrightarrow \text{arg minimal}_c\{ \max_{x\in [0,1]} |x-c| \} (\text{since} [|x-c|+1]^3 \text{is symmetric about $c$} )=0.5 $$
I think this is what you wanted.
