how to find the following series: $$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$ what i attempted was using symmetry like this \begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)} &= 3 \sum_{i,j,n \ge 1} \frac n{n j i (n + j)(n + i)(j + i)} \\ &= 3 \sum_{i,j,n \ge 1} \frac1{i j (n + i)(n + j)(i + j)}. \end{align*} but i don't know how to proceed. also numerically evaluated $$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}\approx 2.47$$ which is roughly $$\frac32\sum_{n=1}^\infty \frac{H_n^2}{n^3}\approx2.47791418905674793594$$ so is $$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}=\frac32\sum_{n=1}^\infty\frac{H_n^2}{n^3}?$$ how to prove this?