By Cauchy-Schwarz,
\[(a^2 + b^2 + c^2 + d^2)(1 + 1 + 1 + 1) \ge (a + b + c + d)^2.\]Thus, $(16 - e^2)(4) \ge (8 - e)^2.$  This simplifies to $16e - 5e^2 \ge 0,$ or $e(16 - 5e) \ge 0.$  Therefore, $e \le \frac{16}{5}.$

Equality occurs when $a = b = c = d = \frac{6}{5}$ and $e = \frac{16}{5},$ so the maximum value of $e$ is $\boxed{\frac{16}{5}}.$
Question:
Given that $a,$ $b,$ $c,$ $d,$ $e$ are real numbers such that
\begin{align*}
a + b + c + d + e &= 8, \\
a^2 + b^2 + c^2 + d^2 + e^2 &= 16.
\end{align*}Determine the maximum value of $e.$
Answer:
Let's think step by step.

user
\frac{16}{5}
Given that $a,$ $b,$ $c,$ $d,$ $e$ are real numbers such that
\begin{align*}
a + b + c + d + e &= 8, \\
a^2 + b^2 + c^2 + d^2 + e^2 &= 16.
\end{align*}Determine the maximum value of $e.$
test