It is well known (the nine-greater-than-eight inequality) that
$$9(a_j + a_k)(a_k + a_i)(a_i + a_j) \geq 8(a_i + a_j + a_k)(a_j a_k + a_k a_i + a_i a_j).$$
Hence the left-hand side is at least
$$\begin{aligned} & \frac{8n}{3} \sum_{i,j,k=1}^{n} \frac{a_i + a_j + a_k}{(a_j + a_k)(a_k + a_i)(a_i + a_j)} \\ &= \frac{4n}{3} \sum_{i,j,k=1}^{n} \left( \frac{1}{(a_k + a_i)(a_i + a_j)} + \frac{1}{(a_i + a_j)(a_j + a_k)} + \frac{1}{(a_j + a_k)(a_k + a_i)} \right) \\ &= 4n \sum_{i,j,k=1}^{n} \frac{1}{(a_k + a_i)(a_i + a_j)}. \end{aligned}$$
Let $S_i = \sum_{\ell=1}^{n} \frac{1}{a_i + a_\ell}$. By the Cauchy-Schwarz inequality we get
$$\begin{aligned} 4n \sum_{i,j,k=1}^{n} \frac{1}{(a_k + a_i)(a_i + a_j)} &= 4n \sum_{i=1}^{n} S_i^2 \\ &\geq \left( \sum_{i=1}^{n} 2S_i \right)^2 \\ &= \left( \sum_{i,l=1}^{n} \frac{2}{a_i + a_l} \right)^2 = \text{R.H.S.,} \end{aligned}$$
and thus the original proposition holds. □