Example 12 Let $a, b, c, d, e$ be real numbers, and $a+b+c+d+e=8, a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16$. Then the maximum value of $e$ is . $\qquad$
Solution: From the given, we have
$$
a+b+c+d=8-e
$$

and $a^{2}+b^{2}+c^{2}+d^{2}=16-e^{2}$.
Let $f(x)=4 x^{2}-2(a+b+c+d) x$
$+\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$
$=(x-a)^{2}+(x-b)^{2}+(x-$
$c)^{2}+(x-d)^{2} \geqslant 0$.
On the other hand, the coefficient of the quadratic term is 4, so $\Delta \leqslant 0$,

i.e., $4(8-e)^{2}-4 \times 4\left(16-e^{2}\right) \leqslant 0$.
Solving this, we get $0 \leqslant e \leqslant \frac{16}{5}$.

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