Setting $d_0 = 1$, we have $j d_j \le d_0 + d_1 + \dots + d_{j-1}$ for any $1 \le j \le n$. Hence
$$\sum_{j=1}^{n} j d_j^2 \le \sum_{j=1}^{n} \sum_{i=0}^{j-1} d_i d_j = \sum_{0 \le i < j \le n} d_i d_j \le \frac{n}{2(n+1)} \left( \sum_{k=0}^{n} d_k \right)^2,$$
where the last part follows from Maclaurin's inequality. Equality holds for $d_1 = d_2 = \dots = d_n = 1$ only.