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

determine the maximum value of $e$.
［Solution］ Using the Cauchy-Schwarz inequality.
Since
$$\begin{aligned}
(a+b+c+d)^{2} & =(a \cdot 1+b \cdot 1+c \cdot 1+d \cdot 1)^{2} \\
& \leqslant\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\left(1^{2}+1^{2}+1^{2}+1^{2}\right) \\
& =4\left(a^{2}+b^{2}+c^{2}+d^{2}\right)
\end{aligned}$$

Also, since
$$\begin{array}{c}
a+b+c+d=8-e \\
a^{2}+b^{2}+c^{2}+d^{2}=16-e^{2}
\end{array}$$

Therefore,
$$(8-e)^{2} \leqslant 4\left(16-e^{2}\right)$$

This gives
$$64-16 e+e^{2} \leqslant 64-4 e^{2} \text { or } 5 e^{2}-16 e \leqslant 0 \text {, }$$
$$0 \leqslant e \leqslant \frac{16}{5}$$

Thus, the maximum value of $e$ is $\frac{16}{5}$.