```markdown
### Proof

Assume that \( \mu > 0 \) (separating hyperplane is nonvertical).

- We can divide the previous equation by \( \mu \) to get:
  \[
  L(x, \tilde{\lambda}/\mu, \tilde{\nu}/\mu) \geq p^\star
  \]
  for all \( x \in \mathcal{D} \).

- Minimizing this inequality over \( x \) produces \( p^\star \leq g(\lambda, \nu) \), where:
  \[
  \lambda = \tilde{\lambda}/\mu,
  \]
  \[
  \nu = \tilde{\nu}/\mu.
  \]

- By weak duality, we have \( g(\lambda, \nu) \leq p^\star \), so in fact \( g(\lambda, \nu) = p^\star \).

This shows that strong duality holds, and that the dual optimum is attained, whenever \( \mu > 0 \). The normal vector has the form \( (\lambda^\star, 1) \) and produces the Lagrange multipliers.

---

*Source: A. d'Aspremont. Convex Optimization M2.*  
*Slide 47/49*
```