Let's solve the problem in **3 steps** as requested: --- **Step 1: Write the Lagrangian** The primal problem is: \[ \begin{align*} \min_{x \geq 0} & \quad \sum_{i=1}^n -\log(x_i + a_i) \\ \text{s.t.} & \quad \sum_{i=1}^n x_i = 1 \end{align*} \] Let \(\lambda\) be the dual variable for the equality constraint. The Lagrangian is: \[ L(x, \lambda) = \sum_{i=1}^n -\log(x_i + a_i) + \lambda \left( \sum_{i=1}^n x_i - 1 \right) \] **Explanation:** We introduce the dual variable \(\lambda\) for the constraint \(\sum x_i = 1\) and write the Lagrangian by combining the objective and constraint. --- **Step 2: Form the Dual Function** The dual function is: \[ g(\lambda) = \inf_{x \geq 0} L(x, \lambda) \] Let's minimize \(L(x, \lambda)\) with respect to \(x_i \geq 0\): \[ L(x, \lambda) = \sum_{i=1}^n \left( -\log(x_i + a_i) + \lambda x_i \right) - \lambda \] Take derivative w.r.t. \(x_i\): \[ \frac{\partial L}{\partial x_i} = -\frac{1}{x_i + a_i} + \lambda \] Set to zero for optimal \(x_i\): \[ -\frac{1}{x_i + a_i} + \lambda = 0 \implies x_i^* = \frac{1}{\lambda} - a_i \] But we need \(x_i^* \geq 0\), so: \[ x_i^* = \max \left\{ 0, \frac{1}{\lambda} - a_i \right\} \] Plug \(x_i^*\) back into the Lagrangian: \[ g(\lambda) = \sum_{i=1}^n \left[ -\log\left( \max\left\{ a_i, \frac{1}{\lambda} \right\} \right) + \lambda \max\left\{ 0, \frac{1}{\lambda} - a_i \right\} \right] - \lambda \] But since \(a_i > 0\), for \(\lambda > 0\) and small enough, \(\frac{1}{\lambda} > a_i\) for all \(i\), so all \(x_i^* = \frac{1}{\lambda} - a_i\), and for larger \(\lambda\), some \(x_i^* = 0\). For dual with one variable, just write for \(\lambda > 0\): \[ x_i^* = \frac{1}{\lambda} - a_i \quad \text{if} \quad \frac{1}{\lambda} > a_i \implies \lambda < \frac{1}{a_i} \] So, for \(\lambda < \min_i \frac{1}{a_i}\), all \(x_i^* > 0\). The constraint \(\sum x_i^* = 1\) gives: \[ \sum_{i=1}^n \left( \frac{1}{\lambda} - a_i \right) = 1 \implies \frac{n}{\lambda} - \sum_{i=1}^n a_i = 1 \implies \frac{n}{\lambda} = 1 + \sum_{i=1}^n a_i \implies \lambda = \frac{n}{1 + \sum_{i=1}^n a_i} \] **Explanation:** We minimize the Lagrangian over \(x\), find the optimal \(x^*\) in terms of \(\lambda\), and plug back into the dual function. --- **Step 3: Write the Dual Problem and Discuss Strong Duality** The dual problem is: \[ \max_{\lambda > 0} g(\lambda) \] Where for feasible \(\lambda\): \[ g(\lambda) = \sum_{i=1}^n \left( -\log \left( \frac{1}{\lambda} \right) + \lambda \left( \frac{1}{\lambda} - a_i \right) \right) - \lambda = -n \log \left( \frac{1}{\lambda} \right) + n - \lambda \sum_{i=1}^n a_i - \lambda \] At the optimal \(\lambda^*\) found above, substitute back: \[ \lambda^* = \frac{n}{1 + \sum_{i=1}^n a_i} \] **Strong duality?** Yes! The objective is convex, the equality constraint is affine, and Slater's condition is satisfied (since \(x_i > 0\) for feasible choices), so **strong duality holds**. --- ### **Final Answers** 1. **Dual problem** (with one dual variable \(\lambda\)) is: \[ \max_{\lambda > 0} g(\lambda) \] where \[ g(\lambda) = \sum_{i=1}^n \left( -\log \left( \frac{1}{\lambda} \right) + \lambda \left( \frac{1}{\lambda} - a_i \right) \right) - \lambda \] and \(\lambda < \min_i \frac{1}{a_i}\). 2. **Strong duality is satisfied** because the problem is convex with affine constraints and Slater's condition holds.