Initial 65. Given a real-coefficient polynomial function $y=a x^{2}+b x+c$, for any $|x| \leqslant 1$, it is known that $|y| \leqslant 1$. Try to find the maximum value of $|a|+|b|+|c|$.
Proof: First, we prove the following auxiliary proposition.
Proposition: Let real numbers $A, B$ satisfy $|A| \leqslant 2, |B| \leqslant 2$. Then $|A+B| + |A-B| \leqslant 4$.
In fact, without loss of generality, let $|A| \geqslant |B|$.
From $A^2 \geqslant B^2$, we have $(A+B)(A-B) \geqslant 0$,
thus $|A+B| + |A-B| = |(A+B) + (A-B)| = |2A| \leqslant 4$.

Now, let's solve the original problem.
When $x=0$, $|y| \leqslant 1$,
i.e., $|c| \leqslant 1$.
For any $|x| \leqslant 1$, we have
\[
|a x^2 + b x| = |y - c| \leqslant |y| + |c| \leqslant 2.
\]

Taking $x=1$, we get $|a + b| \leqslant 2$,
taking $x=-1$, we get $|a - b| \leqslant 2$.
According to the auxiliary proposition, we have
\[
\begin{aligned}
|a| + |b| &= \frac{1}{2}(|2a| + |2b|) \\
&= \frac{1}{2}(|(a+b) + (a-b)| + |(a+b) - (a-b)|) \\
&\leqslant \frac{1}{2} \times 4 = 2.
\end{aligned}
\]

From (1) and (2), we know $|a| + |b| + |c| \leqslant 3$.
Next, we show that $|a| + |b| + |c|$ can achieve the maximum value of 3.
Take $a=2, b=0, c=-1$. In this case, $y = 2x^2 - 1$. It is easy to verify that when $|x| \leqslant 1$, $|y| \leqslant 1$.
Therefore, the maximum value of $|a| + |b| + |c|$ is 3.
(Chen Kuanhong, Nengshi High School, Yueyang County, Hunan Province, 414113)