5. Let $a, b, c \in \mathbb{R}$, for any real number $x$ satisfying $|x| \leq 1$, we have $\left|a x^{2}+b x+c\right| \leq 100$. Then the maximum possible value of $|a|+|b|+|c|$ is
5.300

Analysis: Since $f(1)=a+b+c, f(-1)=a-b+c, f(0)=c$, we have $a=\frac{1}{2}(f(1)+f(-1))-f(0)$, $b=\frac{1}{2}(f(1)-f(-1)), c=f(0)$, and $|f(0)| \leq 100,|f(1)| \leq 100,|f(-1)| \leq 100$, thus
$$
\begin{aligned}
|a|+|b|+|c| & =\left|\frac{1}{2}(f(1)+f(-1))-f(0)\right|+\left|\frac{1}{2}(f(1)-f(-1))\right|+|f(0)| \\
& \leq \frac{1}{2}|f(1)+f(-1)|+|f(0)|+\frac{1}{2}|f(1)-f(-1)|+|f(0)| \\
& \leq \frac{1}{2}(|f(1)+f(-1)|+|f(1)-f(-1)|)+2|f(0)| \leq 300
\end{aligned}
$$

Equality holds if and only if $f(1)=f(-1)=100, f(0)=-100$, i.e., $a x^{2}+b x+c=200 x^{2}-100$