Solution:
If $0 \leq x \leq 1$, then $x^{3 / 2} \leq x$, and equality holds if and only if $x=0$ or $x=1$.

The inequality is true as an equality if all the $a_{k}$'s are zero. Assume that at least one of the numbers $a_{k}$ is non-zero. Set
$$x_{k}=\frac{a_{k}^{2}}{\sum_{j=1}^{n} a_{j}^{2}}$$
Then $0 \leq x_{k} \leq 1$, and by the remark above,
$$\sum_{k=1}^{n}\left(\frac{a_{k}^{2}}{\sum_{j=1}^{n} a_{j}^{2}}\right)^{3 / 2} \leq \sum_{k=1}^{n} \frac{a_{k}^{2}}{\sum_{j=1}^{n} a_{j}^{2}}=1$$
So
$$\sum_{k=1}^{n} a_{k}^{3} \leq\left(\sum_{j=1}^{n} a_{j}^{2}\right)^{3 / 2}$$
which is what was supposed to be proved. For equality, exactly one $x_{k}$ has to be one and the rest have to be zero, which is equivalent to having exactly one of the $a_{k}$'s positive and the rest zero.