Problem: Assume we have an i.i.d data set $\{\bar{x} \} \subset \mathbb{R}^n$, sampled from an $n-$dimensional normal distribution where each dimension is independent, with zero mean and $\sigma^2$ variance (i.e. $P(\bar{x}) = \Pi_k \mathcal{N}(x_k| 0, \sigma^2)$). Prove that $var(\frac{\Vert \bar{x}^i - \bar{x}^j \Vert_2^2}{\mathbb{E}[\Vert \bar{x}^i - \bar{x}^j \Vert_2^2]}) = \frac{n + 2}{n} - 1$.
Attempt at solution: knowing that $\bar{x}^i$ and $\bar{x}^j$ are i.i.d. and normal, we know $\bar{y} = \bar{x}^i - \bar{x}^j$ is also normal, and in fact, $P(y_k) = \mathcal{N}(y_k|0, 2 \sigma^2)$. So I can simplify the target expression to $$var(\frac{\Vert \bar{y} \Vert_2^2}{\mathbb{E}[\Vert \bar{y} \Vert_2^2]}) = var(\frac{\sum_k y_k^2}{\mathbb{E}[\sum_k y_k^2]}) = (\frac{1}{\sum_k \mathbb{E}[y_k^2]})^2 \sum_k var(y_k^2) \\\ = (\frac{1}{\sum_k \mathbb{E}[y_k^2]})^2 \sum_k \mathbb{E}[y_k^4] - \mathbb{E}[y_k^2]^2 \\\ = (\frac{1}{\sum_k \mathbb{E}[y_k^2]})^2 \sum_k 12 \sigma^4 - \mathbb{E}[y_k^2]^2 $$
where in the second line I used a hint for the problem that $\mathbb{E}[y_i^2 y_j^2] = \mathbb{E}[y_i^2] \mathbb{E}[y_j^2]$, and in third line, I used $\mathbb{E}[y_k^4] = 12\sigma^4$ from a hint for the problem. But now I arrive at a wrong result. Any help is appreciated!
Substituting $E[y_k^4] = 12 \sigma^4$ and $E[y_k^2] = 2 \sigma^2$ into your last line yields $$\frac{n(12 - 4) \sigma^4}{(n \cdot 2 \sigma^2)^2} = \frac{2}{n}$$ which is the answer you are looking for.