9. Let $x_{k}=\sqrt{a_{k}}-\sqrt{a_{k+1}}, k=1,2, \cdots, n$. Then we have
$$a_{1}=\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2}, a_{2}=\left(x_{2}+x_{3}+\cdots+x_{n}\right)^{2}, \cdots, a_{n}=x_{n}^{2}$$

Expanding the right-hand side of these equations and adding them, we get
$$\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} k x_{k}^{2}+2 \sum_{1 \leq k<l \leqslant n} k x_{k} x_{l}$$
$x_{k} x_{l}$ appears exactly once in the expansion of each $a_{1}, a_{2}, \cdots, a_{k}$, but does not appear in the expansion of $a_{k+1}$, $a_{k+2}, \cdots a_{n}$, so its coefficient is $k$: Substituting $x_{k}=\sqrt{a_{k}}-\sqrt{a_{k+1}}$ into the right-hand side of the inequality to be proven, and then squaring, we get
$$\left(\sum_{k=1}^{n} \sqrt{k} x_{k}\right)^{2}=\sum_{k=1}^{n} k x_{k}^{2}+2 \sum_{1 \leqslant k<l \leqslant n} \sqrt{k l} x_{k} x_{l}$$

The value of equation (1) is clearly greater than the value of equation (2), so the inequality to be proven holds.