Let $a, b, c, d, e \geq$ such that

$$
a+b+c+d+e=8 \text { and } a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16
$$

What is the maximum value of $e$?
First, let's present Andrea's response on the value of $e$: "it is well known that $e=2.71828 \ldots!"$...

Now, let's move on to the solution. We write the two constraints on $a, b, c, d, e$ as

$$
a+b+c+d=8-e \text{ and } a^{2}+b^{2}+c^{2}+d^{2}=16-e^{2}
$$

Next, we use the Cauchy-Schwarz inequality on $(a, b, c, d)$ and $(1,1,1,1)$:

$$
\begin{aligned}
(a+b+c+d)^{2} & \leq\left(a^{2}+b^{2}+c^{2}+d^{2}\right)(1+1+1+1) \\
(8-e)^{2} & \leq 4\left(16-e^{2}\right) \\
5 e^{2}-16 e & \leq 0
\end{aligned}
$$

A simple study of the quadratic polynomial shows that this is equivalent to $0 \leq e \leq 16 / 5$. By verifying that $e=16 / 5$ with $a=b=c=d=6 / 5$ works.