Q: Minimize x²+y² s.t. x+y ≥ 2. Use KKT. L = x²+y²-λ(x+y-2), λ ≥ 0. ∇L = 0: 2x-λ = 0, 2y-λ = 0: x = y = λ/2. Slackness: λ(x+y-2) = 0. If λ = 0: x = y = 0: violates x+y ≥ 2! So x+y = 2: λ = 2, x = y = 1! Min: 1+1 = 2 ✓ (by symmetry!)!