7. Let us set $c_{k} = A_{k-1} / A_{k}$ for $k = 1, 2, \ldots, n$, where we define $A_{0} = 0$. We observe that $a_{k} / A_{k} = \left( k A_{k} - (k-1) A_{k-1} \right) / A_{k} = k - (k-1) c_{k}$. Now we can write the LHS of the inequality to be proved in terms of $c_{k}$, as follows:
$$\sqrt[n]{\frac{G_{n}}{A_{n}}} = \sqrt[n^{2}]{c_{2} c_{3}^{2} \cdots c_{n}^{n-1}} \text{ and } \frac{g_{n}}{G_{n}} = \sqrt[n]{\prod_{k=1}^{n} \left( k - (k-1) c_{k} \right)}$$
By the AM-GM inequality we have
$$\begin{aligned} n \sqrt[n^{2}]{1^{n(n+1) / 2} c_{2} c_{3}^{2} \ldots c_{n}^{n-1}} & \leq \frac{1}{n} \left( \frac{n(n+1)}{2} + \sum_{k=2}^{n} (k-1) c_{k} \right) \\ & = \frac{n+1}{2} + \frac{1}{n} \sum_{k=1}^{n} (k-1) c_{k} . \end{aligned}$$
Also by the AM-GM inequality, we have
$$\sqrt[n]{\prod_{k=1}^{n} \left( k - (k-1) c_{k} \right)} \leq \frac{n+1}{2} - \frac{1}{n} \sum_{k=1}^{n} (k-1) c_{k}$$
Adding (1) and (2), we obtain the desired inequality. Equality holds if and only if $a_{1} = a_{2} = \cdots = a_{n}$.