一个初等问题

Let $f(x)=ax^2+bx+c$ and $|f(0)|,|f(-1)|,|f(1)|\leq 1$. Then

show that
(1) for any $x\in[-1,1]$, $|f(x)|\leq \frac{5}{4}$.
(2) for any $x\in[-1,1]$, $|cx^2+bx+a|\leq 2$.
(3) for any $x\in[-2,2]$, $|f(x)|\leq 7$.

proof: 
By Lagrange interpolation formula, we have
\begin{align}
f(x)=ax^2+bx+c=\frac{x^2-x}{2}f(-1)+\frac{x^2+x}{2}f(1)+(1-x^2)f(0)
\end{align}
(1) It follows that for any $x\in[-1,1]$,
\begin{align*}
|f(x)|&\leq |\frac{x^2-x}{2}|+|\frac{x^2+x}{2}|+|1-x^2|\\
&=|\frac{x^2-x}{2}-\frac{x^2+x}{2}|+1-x^2\\
&=|x|+1-|x|^2\leq \frac{5}{4},
\end{align*}
where we have used $\frac{x^2-x}{2}\cdot \frac{x^2+x}{2}\leq0$ for $x\in[-1,1]$. In fact, we also can use
\begin{align}
|\frac{x^2-x}{2}|+|\frac{x^2+x}{2}|=\frac{|x|}{2}(|x-1|+|x+1|)=|x|,
\end{align}
since $|x|\leq 1$.
(2)Recall that
\begin{align*}
|cx^2+bx+a|&\leq|c(x^2-1)|+|bx+a+c|\\
&= |f(0)|(1-x^2)|+|\frac{f(1)-f(-1)}{2}x+
\frac{f(1)+f(-1)}{2}|\\
&\leq |f(0)|(1-x^2)|+\frac{1-x}{2}|f(-1)|+\frac{x+1}{2}|f(1)|\\
&\leq 1-x^2+\frac{1-x}{2}+\frac{1+x}{2}\leq 2.
\end{align*}
(3)Recall that ( In fact, By (1), we only need to discuss the case when $1<|x|\leq 2$.)
\begin{align*}
|f(x)|&\leq |\frac{x^2-x}{2}|+|\frac{x^2+x}{2}|+|1-x^2|\\
&=\frac{|x|}{2}(|x-1|+|x+1|)+|1-x^2|\\
&=2|x|+|1-x^2|\leq 7,
\end{align*}
where we have discuss the two case: $|x|\leq 1$ and $1<|x|\leq2$ in the last step.


Note the equality with $"="$ hold with the following functions, respectively.
\begin{align}
-x^2+x+1; \ \ a=2,b=0,c=-1;\ \ 2x^2-1
\end{align}