A problem about quadratic function and inequality Let $f(x) = ax^2 + bx + c(c \neq 0)$. If 
\begin{equation*}
|f(0)| \leqslant 1, |f(1)| \leqslant 1, |f(-1)| \leqslant 1,
\end{equation*} try to show that 
\begin{equation*}
\text{for each } |x| \leqslant 1, \text{ we have }|f(x)| \leqslant \frac{5}{4}.
\end{equation*}
Apparently we can verify this by express $a$, $b$, $c$ in terms of $f(0)$, $f(1)$ and $f(-1)$, 
\begin{equation*}
\left\{ \begin{array}{l} f(0)=c, \\ f(1)=a+b+c, \\ f(-1)=a-b+c. \end{array} \right.  \Longrightarrow  \left\{ \begin{array}{l}a=\frac{f(1)+f(-1)}{2}-f(0), \\ b=\frac{f(1)-f(1)}{2}, \\ c=f(0). \end{array}  \right.
\end{equation*}
I know we can solve it in the above way, but I do wonder if there is any other ideas ??? Is this the only way ???

We have that
$$f(x)=f(0)(1-x)(1+x)+\frac{f(1)}{2}x(1+x)-\frac{f(-1)}{2}x(1-x)$$
then for $x\in [-1,1]$,
$$|f(x)|\leq |f(0)||1-x^2|+\frac{|f(1)|}{2}|x(1+x)|+\frac{|f(-1)|}{2}|x(1-x)|\\
\leq |1-x^2|+\frac{1}{2}|x(1+x)|+\frac{1}{2}|x(1-x)|\\
=(1-x^2)+\frac{1}{2}|x|(1+x)+\frac{1}{2}|x|(1-x)$$
Now the RHS is an even function such that for $x>0$ is equal to $1+x-x^2$. Its maximum value in $[0,1]$ is $5/4$.
