The maximum of $|x^3 + ax^2 + bx + c|$ on $[-1, 1]$ is at least $1/4$ Let $f(x)=x^{3}+ax^{2}+bx+c$ with a, b, c real.
Show that
$$\frac{1}4 \le \max_{-1 \le x \le 1\hspace{2mm}} |f(x)|=M$$
and find all cases where equality occurs.
 A: Note that
$$
\max_{[-1,1]}\,\left|\,x^3-\tfrac34x\,\right|=\tfrac14\tag{1}
$$
Considering the symmetry about $0$ of the domain, we have for any $t\in[0,1]$
$$
\max_{\{t,-t\}}\,\left|\,x^3+ax^2+bx+c\,\right|=\left|\,t^3+bt\,\right|+\left|\,at^2+c\,\right|\tag{2}
$$
Using $(2)$, it is obvious that
$$
M_b=\max_{[-1,1]}\,\left|\,x^3+bx\,\right|\le\max_{[-1,1]}\,\left|\,x^3+ax^2+bx+c\,\right|\tag{3}
$$
It is straightforward to compute
$$
M_b=\left\{\begin{array}{}
2(-b/3)^{3/2}&\text{if }b\in\left[-3,-\tfrac34\right]\\
|\,1+b\,|&\text{otherwise}
\end{array}\right.\tag{4}
$$
and $M_b$ reaches a minimum of $\frac14$ only at $b=-\frac34$. For any other value of $b$, $(3)$ says that
$$
\max_{[-1,1]}\,\left|\,x^3+ax^2+bx+c\,\right|\ge M_b>\tfrac14\tag{5}
$$
Setting $b=-\frac34$ and $t=\frac12$ in $(2)$ yields
$$
\max_{[-1,1]}\,\left|\,x^3+ax^2-\tfrac34x+c\,\right|\ge\tfrac14+\left|\,\tfrac14a+c\,\right|\tag{6}
$$
and this can be $\frac14$ only if $c=-\frac a4$.
At $|x|=\frac12$,
$$
\left|\,x^3+ax^2-\tfrac34x-\tfrac a4\,\right|=\tfrac14\tag{7}
$$
However, at $x=\pm\frac12$, the derivative of $x^3+ax^2-\frac34x-\frac a4$ is $\pm a$. Therefore, the maximum of $\left|\,x^3+ax^2-\tfrac34x-\tfrac a4\,\right|$ will be greater than $\frac14$ unless $a=0$.
Thus,
$$
\max_{[-1,1]}\,\left|\,x^3+ax^2+bx+c\,\right|\ge\tfrac14\tag{8}
$$
where equality holds only for $x^3-\frac34x$.
