Let $T$ be a $n\times J$ matrix with non negative elements, $\mathbf{q,α} \in \Bbb{R}_+^n$ with $α_{i}<1$ for all $i$ and $\mathbf{k}\in\Bbb{R}_+^J$ (with $k_1>0$). Consider the problem
$$\begin{matrix}\min_{\{q_i\}} \sum_{i=1}^n q_{i}^{\frac{1}{α_{i}}}T_{i1} & \\s.t.\sum_{i=1}^n q_{i}^{\frac{1}{α_{i}}}T_{ij}=\frac{k_{j}}{k_1}\sum_{i=1}^n q_{i}^{\frac{1}{α_{i}}}T_{i1} & j=2,\dots,J\\\sum_{i=1}^n q_{i}=1\\ 0≤q_{i}≤1 & i=1,...,n \end{matrix}.$$ Let $\mathbf{q}^*$ be the $\arg \max$ of this problem and define $k_1^*=\sum_{i=1}^n (q_{i}^*)^{\frac{1}{α_{i}}}T_{i1}$ and $k_j^*=r_j k_1$ (for $j=2,\dots,J$). $$\begin{matrix}\max_{\{q_i\}} \sum_{i=1}^n q_{i} & \\s.t.\sum_{i=1}^n q_{i}^{\frac{1}{α_{i}}}T_{ij}\leq k_{j}^* & j=1,\dots,J\\0≤q_{i}≤1 & i=1,...,n \end{matrix}$$
If $\mathbf{k^*}$ is such that the $J$ constraints are satiated, is this second problem the dual of the first? I think they are, but I'd like a second opinion.