. Find all $n$-tuples of strictly positive real numbers $\left(a_{1}, \ldots, a_{n}\right)$ such that:

$$
\sum_{i=1}^{n} a_{i}=96 \quad ; \quad \sum_{i=1}^{n} a_{i}^{2}=144 \quad ; \quad \sum_{i=1}^{n} a_{i}^{3}=216
$$
. According to the Cauchy-Schwarz inequality, we have

$$
\left(a_{1}+\cdots+a_{n}\right)\left(a_{1}^{3}+\cdots+a_{n}^{3}\right) \geqslant\left(a_{1}^{2}+\cdots+a_{n}^{2}\right)^{2}
$$

with equality if and only if $\left(a_{1}, \ldots, a_{n}\right)$ and $\left(a_{1}^{3}, \ldots, a_{n}^{3}\right)$ are proportional. Here, $96 \times 216=144^{2}$: we are therefore in the case of equality. This shows that $a_{1}=\cdots=a_{n}=a$. It follows that $n a=96$ and $n a^{2}=144$, hence $a=3 / 2$, and $n=64$.