Let $a_0 = 1$, yielding $\sum_{i = 0}^{n} a_i = n^n$. Notice that adding $n$ to both sides of the inequality yields: $\sum_{i = 1}^{n}(\frac{\sum_{j=0}^{i}a_j^2}{\sum_{j=0}^{i-1}a_j^2}) > n\frac{n^2}{(n+1)^{\frac{1}{n}}}.$ Taking AM - GM yields: $\sum_{i = 1}^{n}(\frac{\sum_{j=0}^{i}a_i^2}{\sum_{j=0}^{i-1}a_i^2}) \geq n(\sum_{i = 0}^{n} a_i^2)^{\frac{1}{n}}.$ After Cauchy, we obtain: $n(\sum_{i = 0}^{n} a_i^2)^{\frac{1}{n}} = \frac{n}{(n+1)^{\frac{1}{n}}}((n+1)\sum_{i = 0}^{n} a_i^2))^{\frac{1}{n}} \geq \frac{n}{(n+1)^{\frac{1}{n}}} (\sum_{i = 0}^{n} a_i = n^n)^{\frac{2}{n}}$. Notice that the equality case is only satisfied when $1 = a_1 = a_2... a_n$, which is false for $n \geq 2$.