Test $\mathbf{A}$ Given that $a, b, c, d, e$ are real numbers satisfying
$$
\begin{array}{c}
a+b+c+d+e=8, \\
a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16
\end{array}
$$

determine the maximum value of $e$.
Solve for eliminating $a$ and rearranging into a quadratic equation about $b$
$$
2 b^{2}-2(8-c-d-e) b+(8-c-d-e)^{2}+c^{2}+d^{2}+e^{2}-16=0 \text {. }
$$

Since $b \in \mathbf{R}$, its discriminant
$$
\begin{aligned}
\Delta_{1} & =4(8-c-d-e)^{2}-8\left[(8-c-d-e)^{2}+c^{2}+d^{2}+e^{2}-16\right] \\
& \geqslant 0 .
\end{aligned}
$$

Consider it as a quadratic inequality in $c$
$$
3 c^{2}-2(8-d-e) c+(8-d-e)^{2}-2\left(16-d^{2}-e^{2}\right) \leqslant 0 .
$$

Since $c \in \mathbf{R}$, its discriminant
$$
\Delta_{2}=4(8-d-e)^{2}-12\left[(8-d-e)^{2}-2\left(16-d^{2}-e^{2}\right)\right] \geqslant 0 .
$$

Consider it as a quadratic inequality in $d$
$$
4 d^{2}-2(8-e) d+(8-e)^{2}-3\left(16-e^{2}\right) \leqslant 0 .
$$

Since $d \in \mathbf{R}$, we have
$$
\begin{array}{l}
\Delta_{3}=4(8-e)^{2}-16\left[(8-e)^{2}-3\left(16-e^{2}\right)\right] \geqslant 0 \Rightarrow 5 e^{2}-16 e \leqslant 0 \\
\Rightarrow e(5 e-16) \leqslant 0 \Rightarrow 0 \leqslant e \leqslant \frac{16}{5},
\end{array}
$$

Thus,
$$
\max e=\frac{16}{5} .
$$