Let $x\in\mathbb{R}^n_{++}$ be an optimization variable and $\alpha\in\mathbb{R}_{++}^n$ be an $n$-dimensional vector, and $\bar{x}\in\mathbb{R}$ fixed. For any $\varepsilon \geq0$, formulate the problem:
\begin{equation} \begin{array}{ll} \operatorname{minimize} & \max_{i}- \log \left(\frac{\alpha_{i}}{x_{i}+\varepsilon}\right) \\ \text { subject to } & x \succeq 0, \quad \mathbf{1}^{T} x=\bar{x} \end{array} \end{equation} It can be easily observed that the problem is convex since the objective is the maximum of convex functions, and thus convex, and the constraints are affine.
We may reformulate the problem through the auxiliary variable $z$, as follow:
\begin{equation} \begin{array}{ll} \operatorname{minimize} & z \\ \text { subject to } & x \succeq 0, \\& \mathbf{1}^{T} x=\bar{x} \\ & z\geq \log \left(\frac{\alpha_{i}}{x_{i}+\varepsilon}\right), \text{for } i=1,\dots,n \end{array} \end{equation}
The Lagrangian associated with this problem is:
$$ L(x,z,\mu,\lambda,\rho) = z + \mu(\mathbf{1}^{T} x- \bar{x}) -\sum_{i=1}^n\lambda_ix_i-\sum_{i=1} ^n\rho_i\left(\log \left(\frac{\alpha_{i}}{x_{i}+\varepsilon}\right)-z\right).$$ The stationarity conditions impose:
$$(i) \;\frac{dL}{dx_i} = \mu-\lambda_i-\frac{\rho_i}{x_i+\varepsilon} = 0 \Rightarrow \mu = \lambda_i + \frac{\rho_i}{x_i+\varepsilon}, \text{ for } i = 1,\dots,n$$ $$(ii) \;\frac{dL}{dz} = 1-\sum_{i=1}^n\rho_i \Rightarrow \sum_{i=1}^n\rho_i = 1$$
The complementary slackness conditions impose: $$ (iii)\;\lambda_ix_i = 0 \\ (iv)\;\rho_i\left(\log \left(\frac{\alpha_{i}}{x_{i}+\varepsilon}\right)-z\right) = 0\\ (v)\;\mu(\mathbf{1}^{T} x- \bar{x}) = 0 $$
We must also impose primal feasibility, i.e. $$\sum_{i}x_i = \bar{x},$$ and dual feasibility conditions, that is $$\lambda\succeq 0, \rho\succeq 0, \mu\geq 0.$$
How it is possible to proceed from here? I tried to solve this system of the equation but it's not clear to me how to do it.